Mixing
Theorem 11.33 rudin
$begingroup$
Please I have a slight confusion with the notation used by Rudin in the following proof.
11.33 Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and
$$ int_a^b f dx = mathscr{R} int_a^b f dx $$
Where $mathscr{R} int$ denotes the Riemann integral, while $int$ denotes the Lebesgue integral.
Proof Suppose $f$ is bounded. Then there exists a sequence ${P_k}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and
$$ lim_{ktoinfty} L(P_k,f) = mathscr{R}underline{int_a^b} f dx, quad lim_{ktoinfty} U(P_k,f) = mathscr{R}overline{int_a^b} f dx.
$$
Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k={a=x_0<x_1<dots<x_n=b},$ these are defined as,
$$ L(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})m_i, quad U(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})M_i,$$
where $M_i = sup_{xin[x_{i-1},x_i]} f(x)$ and $m_i = inf_{xin[x_{i-1},x_i]} f(x).$
We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x in(x_{i-1},x_i],$ $1leq i leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $xin [a,b],$
$$ L(P_k,f) = int_a^b L_k dx, quad U(P_k,f) = int_a^b U_k dx, $$
and $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$
There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x in [a,b],$
$$ L(x) leq f(x) leq U(x), $$
and by the monotone convergence theorem,
$$
int_a^b L(x) dx = mathscr{R} underline{int_a^b} f dx, quad int_a^b U(x) dx = mathscr{R} overline{int_a^b} f dx.
$$
Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].
Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.
I do understand the idea of the proof but the definition of the partition $P_k={a=x_0<x_1<dots<x_n=b}$. Shouldn't the end point depend on $k$ and not $n$?
Edit: By the explanation of @Ben W, how then is $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$ obtained.
lebesgue-integral riemann-integration
edited Jan 1 at 21:41
amWhy
192k28225439
asked Jan 1 at 21:00
stackuserstackuser
1113
$endgroup$
add a comment |
$begingroup$
Please I have a slight confusion with the notation used by Rudin in the following proof.
11.33 Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and
$$ int_a^b f dx = mathscr{R} int_a^b f dx $$
Where $mathscr{R} int$ denotes the Riemann integral, while $int$ denotes the Lebesgue integral.
Proof Suppose $f$ is bounded. Then there exists a sequence ${P_k}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and
$$ lim_{ktoinfty} L(P_k,f) = mathscr{R}underline{int_a^b} f dx, quad lim_{ktoinfty} U(P_k,f) = mathscr{R}overline{int_a^b} f dx.
$$
Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k={a=x_0<x_1<dots<x_n=b},$ these are defined as,
$$ L(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})m_i, quad U(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})M_i,$$
where $M_i = sup_{xin[x_{i-1},x_i]} f(x)$ and $m_i = inf_{xin[x_{i-1},x_i]} f(x).$
We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x in(x_{i-1},x_i],$ $1leq i leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $xin [a,b],$
$$ L(P_k,f) = int_a^b L_k dx, quad U(P_k,f) = int_a^b U_k dx, $$
and $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$
There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x in [a,b],$
$$ L(x) leq f(x) leq U(x), $$
and by the monotone convergence theorem,
$$
int_a^b L(x) dx = mathscr{R} underline{int_a^b} f dx, quad int_a^b U(x) dx = mathscr{R} overline{int_a^b} f dx.
$$
Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].
Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.
I do understand the idea of the proof but the definition of the partition $P_k={a=x_0<x_1<dots<x_n=b}$. Shouldn't the end point depend on $k$ and not $n$?
Edit: By the explanation of @Ben W, how then is $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$ obtained.
lebesgue-integral riemann-integration
edited Jan 1 at 21:41
amWhy
192k28225439
asked Jan 1 at 21:00
stackuserstackuser
1113
$endgroup$
add a comment |
$begingroup$
Please I have a slight confusion with the notation used by Rudin in the following proof.
11.33 Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and
$$ int_a^b f dx = mathscr{R} int_a^b f dx $$
Where $mathscr{R} int$ denotes the Riemann integral, while $int$ denotes the Lebesgue integral.
Proof Suppose $f$ is bounded. Then there exists a sequence ${P_k}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and
$$ lim_{ktoinfty} L(P_k,f) = mathscr{R}underline{int_a^b} f dx, quad lim_{ktoinfty} U(P_k,f) = mathscr{R}overline{int_a^b} f dx.
$$
Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k={a=x_0<x_1<dots<x_n=b},$ these are defined as,
$$ L(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})m_i, quad U(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})M_i,$$
where $M_i = sup_{xin[x_{i-1},x_i]} f(x)$ and $m_i = inf_{xin[x_{i-1},x_i]} f(x).$
We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x in(x_{i-1},x_i],$ $1leq i leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $xin [a,b],$
$$ L(P_k,f) = int_a^b L_k dx, quad U(P_k,f) = int_a^b U_k dx, $$
and $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$
There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x in [a,b],$
$$ L(x) leq f(x) leq U(x), $$
and by the monotone convergence theorem,
$$
int_a^b L(x) dx = mathscr{R} underline{int_a^b} f dx, quad int_a^b U(x) dx = mathscr{R} overline{int_a^b} f dx.
$$
Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].
Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.
I do understand the idea of the proof but the definition of the partition $P_k={a=x_0<x_1<dots<x_n=b}$. Shouldn't the end point depend on $k$ and not $n$?
Edit: By the explanation of @Ben W, how then is $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$ obtained.
lebesgue-integral riemann-integration
edited Jan 1 at 21:41
amWhy
192k28225439
asked Jan 1 at 21:00
stackuserstackuser
1113
$endgroup$
Please I have a slight confusion with the notation used by Rudin in the following proof.
11.33 Theorem. If $f$ is Riemann integrable on $[a,b],$ then $f$ is Lebesgue integrable on $[a,b]$ with respect to the Lebesgue measure $m$ and
$$ int_a^b f dx = mathscr{R} int_a^b f dx $$
Where $mathscr{R} int$ denotes the Riemann integral, while $int$ denotes the Lebesgue integral.
Proof Suppose $f$ is bounded. Then there exists a sequence ${P_k}$ of partitions of $[a,b]$ such that $P_{k+1}$ is a refinement of $P_k$ for each $k$ and
$$ lim_{ktoinfty} L(P_k,f) = mathscr{R}underline{int_a^b} f dx, quad lim_{ktoinfty} U(P_k,f) = mathscr{R}overline{int_a^b} f dx.
$$
Where $L(P_k), U(P_k)$ are the upper and lower sums respectively. If $P_k={a=x_0<x_1<dots<x_n=b},$ these are defined as,
$$ L(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})m_i, quad U(P_k,f) = sum_{i=1}^n (x_i-x_{i-1})M_i,$$
where $M_i = sup_{xin[x_{i-1},x_i]} f(x)$ and $m_i = inf_{xin[x_{i-1},x_i]} f(x).$
We then define functions $U,L$ as $U_k(a)=L_k(a)=f(a)$ and for each $x in(x_{i-1},x_i],$ $1leq i leq n,$ $U_k(x)=M_i$ and $L_k(x)=m_i.$ Then for all $xin [a,b],$
$$ L(P_k,f) = int_a^b L_k dx, quad U(P_k,f) = int_a^b U_k dx, $$
and $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$
There the sequence of functions $L_k, U_k$ converge point-wise on $[a,b],$ so let $L, U$ be the limit functions respectively. Then $L$ and $U$ are bounded measurable functions on $[a,b]$ and for any $x in [a,b],$
$$ L(x) leq f(x) leq U(x), $$
and by the monotone convergence theorem,
$$
int_a^b L(x) dx = mathscr{R} underline{int_a^b} f dx, quad int_a^b U(x) dx = mathscr{R} overline{int_a^b} f dx.
$$
Since $f$ is Riemann integrable, the upper and lower Riemann integrals are equal. Since $L(x) leq U(x),$ it follows that $L(x) = U(x)$ almost everywhere on [a,b].
Then $L(x) = f(x) = U(x)$ almost everywhere on $[a,b],$ so $f$ is measurable and the result follows.
I do understand the idea of the proof but the definition of the partition $P_k={a=x_0<x_1<dots<x_n=b}$. Shouldn't the end point depend on $k$ and not $n$?
Edit: By the explanation of @Ben W, how then is $$L_1(x) leq L_2(x) leq dots leq f(x) leq dots leq U_2(x) leq U_1(x). $$ obtained.
lebesgue-integral riemann-integration
lebesgue-integral riemann-integration
edited Jan 1 at 21:41
amWhy
192k28225439
asked Jan 1 at 21:00
stackuserstackuser
1113
edited Jan 1 at 21:41
amWhy
192k28225439
asked Jan 1 at 21:00
stackuserstackuser
1113
edited Jan 1 at 21:41
amWhy
192k28225439
edited Jan 1 at 21:41
amWhy
192k28225439
edited Jan 1 at 21:41
amWhy
192k28225439
192k28225439
asked Jan 1 at 21:00
stackuserstackuser
1113
asked Jan 1 at 21:00
stackuserstackuser
1113
asked Jan 1 at 21:00
stackuserstackuser
1113
1113
add a comment |
add a comment |
1 Answer
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For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,140615
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$begingroup$
For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,140615
$endgroup$
add a comment |
$begingroup$
For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,140615
$endgroup$
add a comment |
$begingroup$
For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,140615
$endgroup$
For each $k$ there is $n(k)$ such that
$$P_k={x_0<x_1<cdots<x_{n(k)}}$$
However, it is more convenient just to write $n$ instead of $n(k)$.
answered Jan 1 at 21:03
Ben WBen W
2,140615
answered Jan 1 at 21:03
Ben WBen W
2,140615
answered Jan 1 at 21:03
Ben WBen W
2,140615
answered Jan 1 at 21:03
Ben WBen W
2,140615
2,140615
add a comment |
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