Mixing
Proving $int_{a}^{b}( f+g) = int_{a}^{b} f + int_{a}^{b} g$
$begingroup$
My textbook does the proof in a different way. Here's my attempt:
Since $f,g$ is integrable on $[a,b]$ then there is a family of partitions ${P_n}$ and ${Q_n}$ such that
$$lim_{ntoinfty} L(P_{n},f) = lim_{ntoinfty} U(P_{n}, f)$$
$$lim_{ntoinfty} L(Q_{n},g) = lim_{ntoinfty} U(Q_{n}, g)$$
Let $R_{n} = P_{n} cup Q_{n}$. Since $R_{n}$ is a refinement of $P_{n}$ the following holds :
$$L(P_{n},f) le L(R_{n},f) le U(R_{n}, f) le U(P_{n}, f)$$. By the squeeze theorem, we have that
$$lim_{ntoinfty} L(R_{n},f) = lim_{ntoinfty} U(R_{n}, f)$$
And similarly
$$lim_{ntoinfty} L(R_{n},g) = lim_{ntoinfty} U(R_{n}, g)$$
We note that
$$L(R_n , f) + L(R_n , g) le L(R_n , f+g) le U(R_n , f+g) le U(R_n , f) + U(R_n , g)$$
By the squeeze theorem, we have that $lim_{nto infty} L(R_n , f+g) = lim_{nto infty} U(R_n , f+g) = lim_{nto infty} L(R_n , f) +lim_{nto infty} L(R_n , g)$. Hence it follows that $int_{a}^{b}( f+g) = int_{a}^{b} f + int_{a}^{b} g$
Is this proof correct? Here's the textbook proof:
real-analysis proof-verification riemann-integration
asked Jan 5 at 9:28
Ashish KAshish K
846613
$endgroup$
add a comment |
$begingroup$
My textbook does the proof in a different way. Here's my attempt:
Since $f,g$ is integrable on $[a,b]$ then there is a family of partitions ${P_n}$ and ${Q_n}$ such that
$$lim_{ntoinfty} L(P_{n},f) = lim_{ntoinfty} U(P_{n}, f)$$
$$lim_{ntoinfty} L(Q_{n},g) = lim_{ntoinfty} U(Q_{n}, g)$$
Let $R_{n} = P_{n} cup Q_{n}$. Since $R_{n}$ is a refinement of $P_{n}$ the following holds :
$$L(P_{n},f) le L(R_{n},f) le U(R_{n}, f) le U(P_{n}, f)$$. By the squeeze theorem, we have that
$$lim_{ntoinfty} L(R_{n},f) = lim_{ntoinfty} U(R_{n}, f)$$
And similarly
$$lim_{ntoinfty} L(R_{n},g) = lim_{ntoinfty} U(R_{n}, g)$$
We note that
$$L(R_n , f) + L(R_n , g) le L(R_n , f+g) le U(R_n , f+g) le U(R_n , f) + U(R_n , g)$$
By the squeeze theorem, we have that $lim_{nto infty} L(R_n , f+g) = lim_{nto infty} U(R_n , f+g) = lim_{nto infty} L(R_n , f) +lim_{nto infty} L(R_n , g)$. Hence it follows that $int_{a}^{b}( f+g) = int_{a}^{b} f + int_{a}^{b} g$
Is this proof correct? Here's the textbook proof:
real-analysis proof-verification riemann-integration
asked Jan 5 at 9:28
Ashish KAshish K
846613
$endgroup$
1
$begingroup$
I would say it is.
$endgroup$
– idriskameni
Jan 5 at 9:29
$begingroup$
Minor quibble: your proof is is fine once you prove that reducing the collection of $arbitrary$ partitions to your system ${ P _n}$ is valid, which amounts to proving that if $f$ is Riemann integrable, then for all $epsilon>0$ there is a $delta>0$ such that whenever $P$ is a partition, with mesh $Delta x<delta$, then $|int f-U(P,f)|<epsilon.$
$endgroup$
– Matematleta
Jan 5 at 17:35
$begingroup$
@Matematleta Hey I am not sure if your comment makes much sense to me (since this is my first course in analysis) but isn't it equivalent that there is a sequence of partitions ${P_n}$ such that $lim L(P_n , f) = lim U(P_n , f)$ to saying $f$ is integrable on $[a,b]$?
$endgroup$
– Ashish K
Jan 5 at 23:44
add a comment |
$begingroup$
My textbook does the proof in a different way. Here's my attempt:
Since $f,g$ is integrable on $[a,b]$ then there is a family of partitions ${P_n}$ and ${Q_n}$ such that
$$lim_{ntoinfty} L(P_{n},f) = lim_{ntoinfty} U(P_{n}, f)$$
$$lim_{ntoinfty} L(Q_{n},g) = lim_{ntoinfty} U(Q_{n}, g)$$
Let $R_{n} = P_{n} cup Q_{n}$. Since $R_{n}$ is a refinement of $P_{n}$ the following holds :
$$L(P_{n},f) le L(R_{n},f) le U(R_{n}, f) le U(P_{n}, f)$$. By the squeeze theorem, we have that
$$lim_{ntoinfty} L(R_{n},f) = lim_{ntoinfty} U(R_{n}, f)$$
And similarly
$$lim_{ntoinfty} L(R_{n},g) = lim_{ntoinfty} U(R_{n}, g)$$
We note that
$$L(R_n , f) + L(R_n , g) le L(R_n , f+g) le U(R_n , f+g) le U(R_n , f) + U(R_n , g)$$
By the squeeze theorem, we have that $lim_{nto infty} L(R_n , f+g) = lim_{nto infty} U(R_n , f+g) = lim_{nto infty} L(R_n , f) +lim_{nto infty} L(R_n , g)$. Hence it follows that $int_{a}^{b}( f+g) = int_{a}^{b} f + int_{a}^{b} g$
Is this proof correct? Here's the textbook proof:
real-analysis proof-verification riemann-integration
asked Jan 5 at 9:28
Ashish KAshish K
846613
$endgroup$
My textbook does the proof in a different way. Here's my attempt:
Since $f,g$ is integrable on $[a,b]$ then there is a family of partitions ${P_n}$ and ${Q_n}$ such that
$$lim_{ntoinfty} L(P_{n},f) = lim_{ntoinfty} U(P_{n}, f)$$
$$lim_{ntoinfty} L(Q_{n},g) = lim_{ntoinfty} U(Q_{n}, g)$$
Let $R_{n} = P_{n} cup Q_{n}$. Since $R_{n}$ is a refinement of $P_{n}$ the following holds :
$$L(P_{n},f) le L(R_{n},f) le U(R_{n}, f) le U(P_{n}, f)$$. By the squeeze theorem, we have that
$$lim_{ntoinfty} L(R_{n},f) = lim_{ntoinfty} U(R_{n}, f)$$
And similarly
$$lim_{ntoinfty} L(R_{n},g) = lim_{ntoinfty} U(R_{n}, g)$$
We note that
$$L(R_n , f) + L(R_n , g) le L(R_n , f+g) le U(R_n , f+g) le U(R_n , f) + U(R_n , g)$$
By the squeeze theorem, we have that $lim_{nto infty} L(R_n , f+g) = lim_{nto infty} U(R_n , f+g) = lim_{nto infty} L(R_n , f) +lim_{nto infty} L(R_n , g)$. Hence it follows that $int_{a}^{b}( f+g) = int_{a}^{b} f + int_{a}^{b} g$
Is this proof correct? Here's the textbook proof:
real-analysis proof-verification riemann-integration
real-analysis proof-verification riemann-integration
asked Jan 5 at 9:28
Ashish KAshish K
846613
asked Jan 5 at 9:28
Ashish KAshish K
846613
edited Jan 5 at 9:49
Ashish K
asked Jan 5 at 9:28
Ashish KAshish K
846613
asked Jan 5 at 9:28
Ashish KAshish K
846613
asked Jan 5 at 9:28
Ashish KAshish K
846613
846613
1
$begingroup$
I would say it is.
$endgroup$
– idriskameni
Jan 5 at 9:29
$begingroup$
Minor quibble: your proof is is fine once you prove that reducing the collection of $arbitrary$ partitions to your system ${ P _n}$ is valid, which amounts to proving that if $f$ is Riemann integrable, then for all $epsilon>0$ there is a $delta>0$ such that whenever $P$ is a partition, with mesh $Delta x<delta$, then $|int f-U(P,f)|<epsilon.$
$endgroup$
– Matematleta
Jan 5 at 17:35
$begingroup$
@Matematleta Hey I am not sure if your comment makes much sense to me (since this is my first course in analysis) but isn't it equivalent that there is a sequence of partitions ${P_n}$ such that $lim L(P_n , f) = lim U(P_n , f)$ to saying $f$ is integrable on $[a,b]$?
$endgroup$
– Ashish K
Jan 5 at 23:44
add a comment |
1
$begingroup$
I would say it is.
$endgroup$
– idriskameni
Jan 5 at 9:29
$begingroup$
Minor quibble: your proof is is fine once you prove that reducing the collection of $arbitrary$ partitions to your system ${ P _n}$ is valid, which amounts to proving that if $f$ is Riemann integrable, then for all $epsilon>0$ there is a $delta>0$ such that whenever $P$ is a partition, with mesh $Delta x<delta$, then $|int f-U(P,f)|<epsilon.$
$endgroup$
– Matematleta
Jan 5 at 17:35
$begingroup$
@Matematleta Hey I am not sure if your comment makes much sense to me (since this is my first course in analysis) but isn't it equivalent that there is a sequence of partitions ${P_n}$ such that $lim L(P_n , f) = lim U(P_n , f)$ to saying $f$ is integrable on $[a,b]$?
$endgroup$
– Ashish K
Jan 5 at 23:44
1
1
$begingroup$
I would say it is.
$endgroup$
– idriskameni
Jan 5 at 9:29
$begingroup$
I would say it is.
$endgroup$
– idriskameni
Jan 5 at 9:29
$begingroup$
Minor quibble: your proof is is fine once you prove that reducing the collection of $arbitrary$ partitions to your system ${ P _n}$ is valid, which amounts to proving that if $f$ is Riemann integrable, then for all $epsilon>0$ there is a $delta>0$ such that whenever $P$ is a partition, with mesh $Delta x<delta$, then $|int f-U(P,f)|<epsilon.$
$endgroup$
– Matematleta
Jan 5 at 17:35
$begingroup$
Minor quibble: your proof is is fine once you prove that reducing the collection of $arbitrary$ partitions to your system ${ P _n}$ is valid, which amounts to proving that if $f$ is Riemann integrable, then for all $epsilon>0$ there is a $delta>0$ such that whenever $P$ is a partition, with mesh $Delta x<delta$, then $|int f-U(P,f)|<epsilon.$
$endgroup$
– Matematleta
Jan 5 at 17:35
$begingroup$
@Matematleta Hey I am not sure if your comment makes much sense to me (since this is my first course in analysis) but isn't it equivalent that there is a sequence of partitions ${P_n}$ such that $lim L(P_n , f) = lim U(P_n , f)$ to saying $f$ is integrable on $[a,b]$?
$endgroup$
– Ashish K
Jan 5 at 23:44
$begingroup$
@Matematleta Hey I am not sure if your comment makes much sense to me (since this is my first course in analysis) but isn't it equivalent that there is a sequence of partitions ${P_n}$ such that $lim L(P_n , f) = lim U(P_n , f)$ to saying $f$ is integrable on $[a,b]$?
$endgroup$
– Ashish K
Jan 5 at 23:44
add a comment |
1 Answer
1
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oldest
votes
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We do not, in general, have that $L(f,P)+L(g,P)=L(f+g,P)$ or that $U(f,P)+U(g,P)=U(f+g,P)$ - the infs and sups can happen in different places between the two functions. When you combine the functions, there needs to be an inequality there. After all, it's possible for the sum of functions to be integrable even if neither of the original functions were. By not addressing this, your argument fails.
Now that the argument has been edited, it works.
answered Jan 5 at 9:36
jmerryjmerry
4,434514
$endgroup$
$begingroup$
I was being silly. Let me fix my proof.
$endgroup$
– Ashish K
Jan 5 at 9:43
$begingroup$
Hey is it okay now?
$endgroup$
– Ashish K
Jan 5 at 9:53
1
$begingroup$
Yes, it works now.
$endgroup$
– jmerry
Jan 5 at 10:18
add a comment |
Your Answer
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$begingroup$
We do not, in general, have that $L(f,P)+L(g,P)=L(f+g,P)$ or that $U(f,P)+U(g,P)=U(f+g,P)$ - the infs and sups can happen in different places between the two functions. When you combine the functions, there needs to be an inequality there. After all, it's possible for the sum of functions to be integrable even if neither of the original functions were. By not addressing this, your argument fails.
Now that the argument has been edited, it works.
answered Jan 5 at 9:36
jmerryjmerry
4,434514
$endgroup$
$begingroup$
I was being silly. Let me fix my proof.
$endgroup$
– Ashish K
Jan 5 at 9:43
$begingroup$
Hey is it okay now?
$endgroup$
– Ashish K
Jan 5 at 9:53
1
$begingroup$
Yes, it works now.
$endgroup$
– jmerry
Jan 5 at 10:18
add a comment |
$begingroup$
We do not, in general, have that $L(f,P)+L(g,P)=L(f+g,P)$ or that $U(f,P)+U(g,P)=U(f+g,P)$ - the infs and sups can happen in different places between the two functions. When you combine the functions, there needs to be an inequality there. After all, it's possible for the sum of functions to be integrable even if neither of the original functions were. By not addressing this, your argument fails.
Now that the argument has been edited, it works.
answered Jan 5 at 9:36
jmerryjmerry
4,434514
$endgroup$
$begingroup$
I was being silly. Let me fix my proof.
$endgroup$
– Ashish K
Jan 5 at 9:43
$begingroup$
Hey is it okay now?
$endgroup$
– Ashish K
Jan 5 at 9:53
1
$begingroup$
Yes, it works now.
$endgroup$
– jmerry
Jan 5 at 10:18
add a comment |
$begingroup$
We do not, in general, have that $L(f,P)+L(g,P)=L(f+g,P)$ or that $U(f,P)+U(g,P)=U(f+g,P)$ - the infs and sups can happen in different places between the two functions. When you combine the functions, there needs to be an inequality there. After all, it's possible for the sum of functions to be integrable even if neither of the original functions were. By not addressing this, your argument fails.
Now that the argument has been edited, it works.
answered Jan 5 at 9:36
jmerryjmerry
4,434514
$endgroup$
We do not, in general, have that $L(f,P)+L(g,P)=L(f+g,P)$ or that $U(f,P)+U(g,P)=U(f+g,P)$ - the infs and sups can happen in different places between the two functions. When you combine the functions, there needs to be an inequality there. After all, it's possible for the sum of functions to be integrable even if neither of the original functions were. By not addressing this, your argument fails.
Now that the argument has been edited, it works.
answered Jan 5 at 9:36
jmerryjmerry
4,434514
edited Jan 5 at 10:18
answered Jan 5 at 9:36
jmerryjmerry
4,434514
answered Jan 5 at 9:36
jmerryjmerry
4,434514
answered Jan 5 at 9:36
jmerryjmerry
4,434514
4,434514
$begingroup$
I was being silly. Let me fix my proof.
$endgroup$
– Ashish K
Jan 5 at 9:43
$begingroup$
Hey is it okay now?
$endgroup$
– Ashish K
Jan 5 at 9:53
1
$begingroup$
Yes, it works now.
$endgroup$
– jmerry
Jan 5 at 10:18
add a comment |
$begingroup$
I was being silly. Let me fix my proof.
$endgroup$
– Ashish K
Jan 5 at 9:43
$begingroup$
Hey is it okay now?
$endgroup$
– Ashish K
Jan 5 at 9:53
1
$begingroup$
Yes, it works now.
$endgroup$
– jmerry
Jan 5 at 10:18
$begingroup$
I was being silly. Let me fix my proof.
$endgroup$
– Ashish K
Jan 5 at 9:43
$begingroup$
I was being silly. Let me fix my proof.
$endgroup$
– Ashish K
Jan 5 at 9:43
$begingroup$
Hey is it okay now?
$endgroup$
– Ashish K
Jan 5 at 9:53
$begingroup$
Hey is it okay now?
$endgroup$
– Ashish K
Jan 5 at 9:53
1
1
$begingroup$
Yes, it works now.
$endgroup$
– jmerry
Jan 5 at 10:18
$begingroup$
Yes, it works now.
$endgroup$
– jmerry
Jan 5 at 10:18
add a comment |
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1
$begingroup$
I would say it is.
$endgroup$
– idriskameni
Jan 5 at 9:29
$begingroup$
Minor quibble: your proof is is fine once you prove that reducing the collection of $arbitrary$ partitions to your system ${ P _n}$ is valid, which amounts to proving that if $f$ is Riemann integrable, then for all $epsilon>0$ there is a $delta>0$ such that whenever $P$ is a partition, with mesh $Delta x<delta$, then $|int f-U(P,f)|<epsilon.$
$endgroup$
– Matematleta
Jan 5 at 17:35
$begingroup$
@Matematleta Hey I am not sure if your comment makes much sense to me (since this is my first course in analysis) but isn't it equivalent that there is a sequence of partitions ${P_n}$ such that $lim L(P_n , f) = lim U(P_n , f)$ to saying $f$ is integrable on $[a,b]$?
$endgroup$
– Ashish K
Jan 5 at 23:44