An example of Lebesgue integration












1












$begingroup$


$X = [0,1], mathfrak{M} - sigma$-algebra of Lebesgue measurable subsets from $X$. Prove that $f(x) = e^xcdot chi_{[0,1] setminus mathbb{Q}} + ln{(1+sin{x})} cdot chi_{[0,1] cap mathbb{Q}}$ is Lebesgue integrable on $X$ and calculate $intlimits_{X}fdmu$



Can you please criticize my solution and indicate any mistakes or inaccuracies in reasoning.



1) The value of the Lebesgue integral does not depend on values on zero measure sets, so we can only consider the exponential part.



2) This function is equal to $f(x) = e^x$ on $[0,1]$ almost everywhere, and the latter function is Riemann integrable so the values of the integrals coincide.



3) So the answer is $e - 1$



Here are the questions I would like to clarify:



1) Is the solution above correct and accurate?



2) Is it correct to say that we can always ignore the sets of zero measure?



3) As far as I understand, calculating the Lebesgue integral this way is only possible when considering non-negative functions?



4) The Lebesgue integral is usually first introduced for simple functions. In this case $f(x)$ is not simple function. How do we formally show that it is Lebesgue integrable without appealing to the fact that the function is equivalent to the Riemann integrable function $e^x$ on $[0,1]$?



Thank you in advance!






measure-theory lebesgue-integral lebesgue-measure







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    $X = [0,1], mathfrak{M} - sigma$-algebra of Lebesgue measurable subsets from $X$. Prove that $f(x) = e^xcdot chi_{[0,1] setminus mathbb{Q}} + ln{(1+sin{x})} cdot chi_{[0,1] cap mathbb{Q}}$ is Lebesgue integrable on $X$ and calculate $intlimits_{X}fdmu$



    Can you please criticize my solution and indicate any mistakes or inaccuracies in reasoning.



    1) The value of the Lebesgue integral does not depend on values on zero measure sets, so we can only consider the exponential part.



    2) This function is equal to $f(x) = e^x$ on $[0,1]$ almost everywhere, and the latter function is Riemann integrable so the values of the integrals coincide.



    3) So the answer is $e - 1$



    Here are the questions I would like to clarify:



    1) Is the solution above correct and accurate?



    2) Is it correct to say that we can always ignore the sets of zero measure?



    3) As far as I understand, calculating the Lebesgue integral this way is only possible when considering non-negative functions?



    4) The Lebesgue integral is usually first introduced for simple functions. In this case $f(x)$ is not simple function. How do we formally show that it is Lebesgue integrable without appealing to the fact that the function is equivalent to the Riemann integrable function $e^x$ on $[0,1]$?



    Thank you in advance!






    measure-theory lebesgue-integral lebesgue-measure







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      $X = [0,1], mathfrak{M} - sigma$-algebra of Lebesgue measurable subsets from $X$. Prove that $f(x) = e^xcdot chi_{[0,1] setminus mathbb{Q}} + ln{(1+sin{x})} cdot chi_{[0,1] cap mathbb{Q}}$ is Lebesgue integrable on $X$ and calculate $intlimits_{X}fdmu$



      Can you please criticize my solution and indicate any mistakes or inaccuracies in reasoning.



      1) The value of the Lebesgue integral does not depend on values on zero measure sets, so we can only consider the exponential part.



      2) This function is equal to $f(x) = e^x$ on $[0,1]$ almost everywhere, and the latter function is Riemann integrable so the values of the integrals coincide.



      3) So the answer is $e - 1$



      Here are the questions I would like to clarify:



      1) Is the solution above correct and accurate?



      2) Is it correct to say that we can always ignore the sets of zero measure?



      3) As far as I understand, calculating the Lebesgue integral this way is only possible when considering non-negative functions?



      4) The Lebesgue integral is usually first introduced for simple functions. In this case $f(x)$ is not simple function. How do we formally show that it is Lebesgue integrable without appealing to the fact that the function is equivalent to the Riemann integrable function $e^x$ on $[0,1]$?



      Thank you in advance!






      measure-theory lebesgue-integral lebesgue-measure







      share|cite|improve this question











      $endgroup$




      $X = [0,1], mathfrak{M} - sigma$-algebra of Lebesgue measurable subsets from $X$. Prove that $f(x) = e^xcdot chi_{[0,1] setminus mathbb{Q}} + ln{(1+sin{x})} cdot chi_{[0,1] cap mathbb{Q}}$ is Lebesgue integrable on $X$ and calculate $intlimits_{X}fdmu$



      Can you please criticize my solution and indicate any mistakes or inaccuracies in reasoning.



      1) The value of the Lebesgue integral does not depend on values on zero measure sets, so we can only consider the exponential part.



      2) This function is equal to $f(x) = e^x$ on $[0,1]$ almost everywhere, and the latter function is Riemann integrable so the values of the integrals coincide.



      3) So the answer is $e - 1$



      Here are the questions I would like to clarify:



      1) Is the solution above correct and accurate?



      2) Is it correct to say that we can always ignore the sets of zero measure?



      3) As far as I understand, calculating the Lebesgue integral this way is only possible when considering non-negative functions?



      4) The Lebesgue integral is usually first introduced for simple functions. In this case $f(x)$ is not simple function. How do we formally show that it is Lebesgue integrable without appealing to the fact that the function is equivalent to the Riemann integrable function $e^x$ on $[0,1]$?



      Thank you in advance!






      measure-theory lebesgue-integral lebesgue-measure




      measure-theory lebesgue-integral lebesgue-measure






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Feb 2 at 16:56







      Don Draper

















      asked Feb 2 at 12:36









      Don DraperDon Draper

      87110




      87110






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Your solution is valid, but if you are raising these questions then perhaps you have not absorbed the basic theory of the Lebesgue integral, and for this you also need to understand the concepts of Lebesgue measurability of sets and functions.



          (2) That we can ignore sets of measure zero follows from a basic proposition (usually stated and proved in standard textbooks). If $f$ is Lebesgue integrable and $f = g$ almost everywhere then $int_X f = int_Xg$. This follows from basic properties of additivity and the fact that integrals over zero measure sets are zero. If $m(E) = 0$, e.g. $E={x : f(x) neq g(x)}$ in this case, then



          $$int_Xf = int_{X setminus E}f + int_Ef = int_{X setminus E}f = int_{X setminus E}g = int_Xg$$



          (3) & (4) Non-negativity is not needed here. For bounded functions on sets of finite measure we can say the function is Lebesgue integrable if



          $$sup_{phi leqslant f} int_X phi = inf_{psi geqslant f}int_X psi,$$



          where $phi, psi$ denote simple functions. In this case the Lebesgue integral of $f$ is the common value of these lower and upper Lebesgue integrals, analogous to lower and upper Darboux integrals.



          We then have the key result that a bounded function $f$ on a finite measure set $X$ is Lebesgue integrable if and only $f$ is measureable. Presumably you should know that any continuous function is Lebesgue measureable. Any Riemann integrable function is also measureable and from this we can show that the Riemann integral over a compact interval coincides with the Lebesgue integral. Either of these results allows you to conclude that $x mapsto e^x$ is Lebesgue integrable over $X = [0,1]$.



          The next step in the development of the Lebesgue integral is to consider functions that may not be bounded and sets $X$ that may not have finite measure. Here is where the integral is first defined for non-negative functions and then later extended by taking integrals of the positive and negative parts.



          Now it is typical of these exercises to ask you to compute the Lebesgue integral. This can usually be accomplished by finding a sequence $f_n$ converging pointwise to $f$ where $int_X f_n$ is known and then appealing to the monotone or dominated convergence theorem to get



          $$int_X f = lim_{n to infty} int_X f_n$$



          When $X$ is a compact interval and $f$ Riemann is integrable, then, if you are averse to simply declaring that the Lebesgue integral is the Riemann integral, you can always take step functions $f_n$ that correspond to lower Darboux sums. Also any measurable function is the limit of a sequence of simple functions (increasing if $f$ is nonnegative).






          share|cite|improve this answer











          $endgroup$














            Your Answer








            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097252%2fan-example-of-lebesgue-integration%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Your solution is valid, but if you are raising these questions then perhaps you have not absorbed the basic theory of the Lebesgue integral, and for this you also need to understand the concepts of Lebesgue measurability of sets and functions.



            (2) That we can ignore sets of measure zero follows from a basic proposition (usually stated and proved in standard textbooks). If $f$ is Lebesgue integrable and $f = g$ almost everywhere then $int_X f = int_Xg$. This follows from basic properties of additivity and the fact that integrals over zero measure sets are zero. If $m(E) = 0$, e.g. $E={x : f(x) neq g(x)}$ in this case, then



            $$int_Xf = int_{X setminus E}f + int_Ef = int_{X setminus E}f = int_{X setminus E}g = int_Xg$$



            (3) & (4) Non-negativity is not needed here. For bounded functions on sets of finite measure we can say the function is Lebesgue integrable if



            $$sup_{phi leqslant f} int_X phi = inf_{psi geqslant f}int_X psi,$$



            where $phi, psi$ denote simple functions. In this case the Lebesgue integral of $f$ is the common value of these lower and upper Lebesgue integrals, analogous to lower and upper Darboux integrals.



            We then have the key result that a bounded function $f$ on a finite measure set $X$ is Lebesgue integrable if and only $f$ is measureable. Presumably you should know that any continuous function is Lebesgue measureable. Any Riemann integrable function is also measureable and from this we can show that the Riemann integral over a compact interval coincides with the Lebesgue integral. Either of these results allows you to conclude that $x mapsto e^x$ is Lebesgue integrable over $X = [0,1]$.



            The next step in the development of the Lebesgue integral is to consider functions that may not be bounded and sets $X$ that may not have finite measure. Here is where the integral is first defined for non-negative functions and then later extended by taking integrals of the positive and negative parts.



            Now it is typical of these exercises to ask you to compute the Lebesgue integral. This can usually be accomplished by finding a sequence $f_n$ converging pointwise to $f$ where $int_X f_n$ is known and then appealing to the monotone or dominated convergence theorem to get



            $$int_X f = lim_{n to infty} int_X f_n$$



            When $X$ is a compact interval and $f$ Riemann is integrable, then, if you are averse to simply declaring that the Lebesgue integral is the Riemann integral, you can always take step functions $f_n$ that correspond to lower Darboux sums. Also any measurable function is the limit of a sequence of simple functions (increasing if $f$ is nonnegative).






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Your solution is valid, but if you are raising these questions then perhaps you have not absorbed the basic theory of the Lebesgue integral, and for this you also need to understand the concepts of Lebesgue measurability of sets and functions.



              (2) That we can ignore sets of measure zero follows from a basic proposition (usually stated and proved in standard textbooks). If $f$ is Lebesgue integrable and $f = g$ almost everywhere then $int_X f = int_Xg$. This follows from basic properties of additivity and the fact that integrals over zero measure sets are zero. If $m(E) = 0$, e.g. $E={x : f(x) neq g(x)}$ in this case, then



              $$int_Xf = int_{X setminus E}f + int_Ef = int_{X setminus E}f = int_{X setminus E}g = int_Xg$$



              (3) & (4) Non-negativity is not needed here. For bounded functions on sets of finite measure we can say the function is Lebesgue integrable if



              $$sup_{phi leqslant f} int_X phi = inf_{psi geqslant f}int_X psi,$$



              where $phi, psi$ denote simple functions. In this case the Lebesgue integral of $f$ is the common value of these lower and upper Lebesgue integrals, analogous to lower and upper Darboux integrals.



              We then have the key result that a bounded function $f$ on a finite measure set $X$ is Lebesgue integrable if and only $f$ is measureable. Presumably you should know that any continuous function is Lebesgue measureable. Any Riemann integrable function is also measureable and from this we can show that the Riemann integral over a compact interval coincides with the Lebesgue integral. Either of these results allows you to conclude that $x mapsto e^x$ is Lebesgue integrable over $X = [0,1]$.



              The next step in the development of the Lebesgue integral is to consider functions that may not be bounded and sets $X$ that may not have finite measure. Here is where the integral is first defined for non-negative functions and then later extended by taking integrals of the positive and negative parts.



              Now it is typical of these exercises to ask you to compute the Lebesgue integral. This can usually be accomplished by finding a sequence $f_n$ converging pointwise to $f$ where $int_X f_n$ is known and then appealing to the monotone or dominated convergence theorem to get



              $$int_X f = lim_{n to infty} int_X f_n$$



              When $X$ is a compact interval and $f$ Riemann is integrable, then, if you are averse to simply declaring that the Lebesgue integral is the Riemann integral, you can always take step functions $f_n$ that correspond to lower Darboux sums. Also any measurable function is the limit of a sequence of simple functions (increasing if $f$ is nonnegative).






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Your solution is valid, but if you are raising these questions then perhaps you have not absorbed the basic theory of the Lebesgue integral, and for this you also need to understand the concepts of Lebesgue measurability of sets and functions.



                (2) That we can ignore sets of measure zero follows from a basic proposition (usually stated and proved in standard textbooks). If $f$ is Lebesgue integrable and $f = g$ almost everywhere then $int_X f = int_Xg$. This follows from basic properties of additivity and the fact that integrals over zero measure sets are zero. If $m(E) = 0$, e.g. $E={x : f(x) neq g(x)}$ in this case, then



                $$int_Xf = int_{X setminus E}f + int_Ef = int_{X setminus E}f = int_{X setminus E}g = int_Xg$$



                (3) & (4) Non-negativity is not needed here. For bounded functions on sets of finite measure we can say the function is Lebesgue integrable if



                $$sup_{phi leqslant f} int_X phi = inf_{psi geqslant f}int_X psi,$$



                where $phi, psi$ denote simple functions. In this case the Lebesgue integral of $f$ is the common value of these lower and upper Lebesgue integrals, analogous to lower and upper Darboux integrals.



                We then have the key result that a bounded function $f$ on a finite measure set $X$ is Lebesgue integrable if and only $f$ is measureable. Presumably you should know that any continuous function is Lebesgue measureable. Any Riemann integrable function is also measureable and from this we can show that the Riemann integral over a compact interval coincides with the Lebesgue integral. Either of these results allows you to conclude that $x mapsto e^x$ is Lebesgue integrable over $X = [0,1]$.



                The next step in the development of the Lebesgue integral is to consider functions that may not be bounded and sets $X$ that may not have finite measure. Here is where the integral is first defined for non-negative functions and then later extended by taking integrals of the positive and negative parts.



                Now it is typical of these exercises to ask you to compute the Lebesgue integral. This can usually be accomplished by finding a sequence $f_n$ converging pointwise to $f$ where $int_X f_n$ is known and then appealing to the monotone or dominated convergence theorem to get



                $$int_X f = lim_{n to infty} int_X f_n$$



                When $X$ is a compact interval and $f$ Riemann is integrable, then, if you are averse to simply declaring that the Lebesgue integral is the Riemann integral, you can always take step functions $f_n$ that correspond to lower Darboux sums. Also any measurable function is the limit of a sequence of simple functions (increasing if $f$ is nonnegative).






                share|cite|improve this answer











                $endgroup$



                Your solution is valid, but if you are raising these questions then perhaps you have not absorbed the basic theory of the Lebesgue integral, and for this you also need to understand the concepts of Lebesgue measurability of sets and functions.



                (2) That we can ignore sets of measure zero follows from a basic proposition (usually stated and proved in standard textbooks). If $f$ is Lebesgue integrable and $f = g$ almost everywhere then $int_X f = int_Xg$. This follows from basic properties of additivity and the fact that integrals over zero measure sets are zero. If $m(E) = 0$, e.g. $E={x : f(x) neq g(x)}$ in this case, then



                $$int_Xf = int_{X setminus E}f + int_Ef = int_{X setminus E}f = int_{X setminus E}g = int_Xg$$



                (3) & (4) Non-negativity is not needed here. For bounded functions on sets of finite measure we can say the function is Lebesgue integrable if



                $$sup_{phi leqslant f} int_X phi = inf_{psi geqslant f}int_X psi,$$



                where $phi, psi$ denote simple functions. In this case the Lebesgue integral of $f$ is the common value of these lower and upper Lebesgue integrals, analogous to lower and upper Darboux integrals.



                We then have the key result that a bounded function $f$ on a finite measure set $X$ is Lebesgue integrable if and only $f$ is measureable. Presumably you should know that any continuous function is Lebesgue measureable. Any Riemann integrable function is also measureable and from this we can show that the Riemann integral over a compact interval coincides with the Lebesgue integral. Either of these results allows you to conclude that $x mapsto e^x$ is Lebesgue integrable over $X = [0,1]$.



                The next step in the development of the Lebesgue integral is to consider functions that may not be bounded and sets $X$ that may not have finite measure. Here is where the integral is first defined for non-negative functions and then later extended by taking integrals of the positive and negative parts.



                Now it is typical of these exercises to ask you to compute the Lebesgue integral. This can usually be accomplished by finding a sequence $f_n$ converging pointwise to $f$ where $int_X f_n$ is known and then appealing to the monotone or dominated convergence theorem to get



                $$int_X f = lim_{n to infty} int_X f_n$$



                When $X$ is a compact interval and $f$ Riemann is integrable, then, if you are averse to simply declaring that the Lebesgue integral is the Riemann integral, you can always take step functions $f_n$ that correspond to lower Darboux sums. Also any measurable function is the limit of a sequence of simple functions (increasing if $f$ is nonnegative).







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Feb 3 at 6:57

























                answered Feb 3 at 6:38









                RRLRRL

                53.7k52574




                53.7k52574






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097252%2fan-example-of-lebesgue-integration%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    MongoDB - Not Authorized To Execute Command

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I have successfully enabled authorization on MongoDB and I have created an account on the admin database and then I created an account for my database called test. The following connection string to connect to my test database works successfully: mongo --host 192.168.17.52 --port 27017 -u user1 -p password --authenticationDatabase test Only problem I have now is, I cannot execute commands such as: show dbs. I get the following error when I try to do so: "errmsg" : "not authorized on admin to execute command { listDatabases: 1.0, lsid: { id: UUID("a1d5bc0d-bc58-485e-b232-270758a89455") }, $db: "admin...
                    Read more

                    How to fix TextFormField cause rebuild widget in Flutter

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 0 I've got the same issue at #9280. I tried all solutions from that topic with/without GlobalKey for Scaffold and Form but the problem wasn't fixed. I tried to build a authentication form with login and signup. But the keyboard doesn't show when I tap the TextFormField at the first time. Then I try to input some letter from physical keyboard, the widget is rebuilt. Like this: login screen Nothing to be log when I run flutter run --verbose, so I logged manually. Every tapping on field, the widget called dispose -> init -> build. [√] Flutter (Channel beta, v1.0.0, on Microsoft Windows [Version 10.0.17763.195], ...
                    Read more

                    in spring boot 2.1 many test slices are not allowed anymore due to multiple @BootstrapWith

                    Image
                    1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
                    Read more