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Prove that a bounded function $f$ is integrable on $[0,1]$ if $$I_0 := \lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n),$$ in which case $\int_0^1f(x)dx$ equals $I_0$.

Refer here. I suspect that this is the same question with answers but I am not sure how to apply it to prove my particular case.

Furthermore, consider the definition that I was given below

Definition: $f : [a,b]\to \mathbb R$ is said to be (Riemann) integrable on $[a,b]$ if and only if $f$ is bounded on $[a,b]$, and for every $\epsilon > 0$ there is a partition $P$ of $[a,b]$ such that $U(f,P) - L(f,P) < \epsilon$.

With this definition in mind, why then did the linked post above need to show that $$\overline{\int}_a^b f \leq U(f,P_N) \leq L(f,P_N) \leq \underline{\int}_a^b f?$$ Would it not just follows from the definition (in my particular case) that if $U=L$ as $n\to \infty$ then there automatically is a partition $P$ such that $U(f,P) - L(f,P) < \epsilon$ is satisfied by definition?

Any clarification is helpful thank you!

Owen Murphy
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Beccause the upper and lower integrals are the infimum and supremum, respectively, of upper and lower sums, it follows that for any $N$ and $M$

$$L(f,P_N) \leqslant \underline{\int}_a^b f \leqslant \overline{\int_a}^bf\leqslant U(f,P_M)$$

Since $I_0 = \lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n)$, for any $\epsilon > 0$ there exists $N$ and $M$ such that

$$I_0 - \epsilon < L(f,P_N) \leqslant U(f, P_M ) < I_0 +\epsilon$$

Thus,

$$I_0 - \epsilon < \underline{\int}_a^b f \leqslant \overline{\int_a}^bf< I_0 +\epsilon$$

Since $\epsilon$ can be arbitrarily close to $0$, it follows that

$$I_0= \underline{\int}_a^b f = \overline{\int_a}^bf$$

This proves both that $f$ is Riemann integrable and that $I_0$ is the value of the integral.

RRL
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    This also uses the fact that given any partitions $P$ and $Q$, we must have $L(f,P) \leqslant U(f,Q)$ which implies that $\underline{\int}_a^b f \leqslant \overline{\int_a}^bf$ (for any bounded $f$ even if it is not Riemann integrable). – RRL Dec 02 '21 at 03:42
  • I see, so Based on your first sentence, in other words, taking the $\inf(U) - \sup(L) < \epsilon$ is essentially equivalent to $U - L < \epsilon$ in proving that an $f$ is Riemann integrable? Forgive me if I am mistaken. – Owen Murphy Dec 02 '21 at 03:50
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    Yes. The usual Riemann criterion is that if we can find a partition such that the the difference between corresponding upper and lower sums is less than $\epsilon$, and this works for any $\epsilon$, then the function is Riemann integrable. Again this is because the upper and lower integrals are squeezed and must differ by less than any $\epsilon$. The proof here is similar except we don’t use upper and lower sums for the same partition. – RRL Dec 02 '21 at 04:07
  • Dear @RRL, thank you for a nice proof. I would be extremely grateful if you could prove the converse statement in an edit, that is, if $f$ is integrable, then $I_0 = \lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n)$. – psie Apr 24 '23 at 12:39
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    @schn: Thanks. It is clearly not true for every partition sequence $(P_n)$ -- for example, a constant sequence. Do you mean prove there exists a sequence where the lower and upper sums converge to $I_0$? Or perhaps with an added condition such as $|P_n| \to 0$. – RRL Apr 24 '23 at 15:31
  • Yes. If I understood the OP's question correctly, you proved "if there is a $P_n$ such that $ I_0 = \lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n)$, then $f$ is integrable and $I_0$ is the value of the integral". So I was wondering if the converse also holds: "if $f$ is integrable and $I_0$ is the value of the integral, then there exists a $P_n$ such that $ I_0 = \lim_{n\to\infty}L(f,P_n) = \lim_{n\to\infty}U(f,P_n)$". I don't know if this statement holds, but I am trying to go from bottom to top in your post. I am stuck on how to obtain the second inequality from the third :-) – psie Apr 24 '23 at 22:00
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    @schn: That converse is true. When I get a chance I’ll show you how to proceed. – RRL Apr 25 '23 at 05:18
  • My attempt so far is the following: if $f$ is integrable and $I_0= \underline{\int}_a^b f = \overline{\int_a}^bf,$ then, for $\epsilon>0$, clearly $I_0 - \epsilon < \underline{\int}_a^b f = \overline{\int_a}^bf< I_0 +\epsilon.$ Next I would like to somehow squeeze in an $L(f,P_n)$ such that $I_0 - \epsilon <L(f,P_n)\leq \underline{\int}_a^b f$ and similarly for $U(f,P_n)$. If such a sequence of partition exists, probably for large enough $n$, then we would be done. But I'm unsure which property justifies such a sequence and which conditions arise on the sequence. – psie Apr 25 '23 at 19:24
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    @schn: Pose this as a new question showing your work as you show here. Comments are really not meant for extensive discussion and the posting of solutions. It is also better to not answer a second question from comments by appending into an existing answer from an earlier question. – RRL Apr 25 '23 at 19:32
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    @schn: I can help you then. – RRL Apr 25 '23 at 19:35
  • Ok, I understand. I asked a question about this implication the other day :-) – psie Apr 25 '23 at 19:42
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    @schn: I added an answer there. – RRL Apr 25 '23 at 19:50