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I'm not sure if the following prove is correct. It seems incomplete. Hope you can help me:

Theorem: Suppose $f:[a,b] \to \mathbb R$ is Riemann integrable. Suppose $g: [a,b] \to \mathbb R$ is a function such that $g(x) = f(x)$ for all except finite many $x \in [a,b]$. Prove that $g$ is Riemann integrable in $[a,b]$. and

$$ \int_{a}^{b} f = \int_{a}^{b} g $$

Prove:
If f is integrable then for each partition $P$ of $[a,b]$ there exist an $\epsilon\gt 0$ such that $$ \\ U(f,[a,b])-L(f,[a,b]) < \epsilon \\\ \implies U(f,[a,b])-L(f,[a,b]) \le U(f,P,[a,b])-L(f,P,[a,b]) \\\ \\\ = \sum_{j=1}^{n} (x_{j}-x_{j-1}) \sup f - \sum_{j=1}^{n} (x_{j}-x_{j-1}) \inf f \\\ $$

since $f(x)=g(x)$ for $x \in [a,b]$

$$ \sum_{j=1}^{n} (x_{j}-x_{j-1}) \sup g - \sum_{j=1}^{n} (x_{j}-x_{j-1}) \inf g \\\ = U(g,[a,b])-L(g,[a,b]) \le U(g,P,[a,b])-L(g,P,[a,b]) \\\ $$

Therefore $g$ is also Riemann integrable and

$$ \int_{a}^{b} f = \int_{a}^{b} g $$

Edit:

Next try:

If e.g. the value of the last point of an interval $[a,b]$ is $g(x)=f(x)+1$ and for the rest of the points $g(x)=f(x)$ then with any partition P

$$ U(h,P,[a,b])=\sum_{i=1}^{n-1}(x_{j}-x_{j-1})\sup_{[x_{j-1},x_{j}]} h + (x_{n}-x_{n-1})\sup_{[x_{n-1},x_{n}]} h \\\ \lim_{n\to\infty} \sum_{i=1}^{n-1}(x_{j}-x_{j-1})\sup_{[x_{j-1},x_{j}]} h + (x_{n}-x_{n-1})\sup_{[x_{n-1},x_{n}]} h = 0 $$

with

$$ h(x) = f(x)-g(x) $$

This can be applied to each subinterval of [a,b].

Does this work?

songwriter93
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  • $f(x) = g(x)$ is not true for finitely many points as you stated in your hypotheses so the sup g in some of the subintervals could be different from that of f, right? – Aria Oct 10 '23 at 19:43
  • I think the idea is that you have to make those subintervals containing those finitely many points arbitrarily small to neutralize their effect. – Aria Oct 10 '23 at 19:44
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    Your proof needs some work. Take $f=0$ and any partition $P$. Now pick some point $x'$ in the interior of one of the partition intervals, say $x' \in [a',b']$ (where $a',b'$ are neighbouring points in the partition) and let $g=0$ everywhere except $g(x') = 1$. Then $L(f,P)=U(f,P)$ but $L(g,P) = 0$, $U(g,P) = b'-a'$. – copper.hat Oct 10 '23 at 19:51
  • @Aria thanks for the hint! So I can understand the "except finite many x" as "a few" points that are not equal to f? – songwriter93 Oct 10 '23 at 19:52
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    It suffices to show that $h=g-f$ is Riemann integrable. The function $h$ is a finite sum of functions $h_k$ each nonzero at a single point. So it suffices to prove that $h_k$ are integrable. The latter should be easy to handle. – Ryszard Szwarc Oct 10 '23 at 19:52
  • @songwriter93, you're welcome. "Except for finitely many points" means that $f(x)$ might not be equal to $g(x)$ for a finite number of x like for 100, 10000 or 1000^1000 points, but even if there are finitely many of them, they can still change the supremum and infimum of g(x) in some subintervals, but fortunately since there are only finite numbers of them, you can make those subintervals small enough to solve the problem. the ideas of the other comments are also very interesting and shorter than mine. – Aria Oct 10 '23 at 19:57
  • Okay these are some good explanations. Thank you all! I'll try to solve it and push you my results. – songwriter93 Oct 10 '23 at 20:00
  • You only need to prove it for a single point of difference. – copper.hat Oct 10 '23 at 20:17
  • @copper.hat Do I have to generalize it after proving it with a single point to make it a complete prove?? – songwriter93 Oct 10 '23 at 20:48
  • Yes indeed, but it is just a case of adding one point at a time. – copper.hat Oct 10 '23 at 20:59

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