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Here is my proof.

Theorem: If $f$ is Riemann integrable on $[a,b],c\in[a,b]$, and $g(x)=f(x)$ except for finitely many points $c_1,\cdots,c_k$ in $[a,b]$, then $g$ is Riemann integrable on $[a,b]$, and $\int_{a}^{b}f(x)dx=\int_{a}^{b}g(x)dx.$

Suppose another function $g$ such that $g(x)=f(x)$ except for finitely many points $c_1,\cdots,c_k$. Since $f$ is Riemann integrable, for any $\epsilon>0$, there exists $\delta=\frac{\epsilon}{4n\sum_{i=1}^{k}|g(c_i)-f(c_i)|}$ such that for any partition $P=(a=x_0,\cdots,x_n=b)$ with $\left\lVert P\right\rVert<\delta$, we have $\left|R(f,P)-\int_{a}^{b}f\right|<\frac{\epsilon}{2}.$ Now, \begin{align*} \left|R(g,P)-\int_{a}^{b}f\right|&=\left|R(g,P)-R(f,P)+R(f,P)-\int_{a}^{b}f\right|\\ &\leq\left|R(g,P)-R(f,P)\right|+\left|R(f,P)-\int_{a}^{b}f\right|\\ &<\left|R(g,P)-R(f,P)\right|+\frac{\epsilon}{2}\\ &<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\\ \end{align*}where \begin{align*} |R(g,P)-R(f,P)|&=\left|\sum_{i=1}^{n}g(x_i*)(x_i-x_{i-1})-\sum_{i=1}^{n}f(x_i*)(x_i-x_{i-1})\right|\\ &=\left|\sum_{i=1}^{n}(g(x_i^*)-f(x_i^*))(x_i-x_{i-1})\right|\\ &<\sum_{i=1}^{k}|g(c_i)-f(c_i)|(2n\delta)\\ &=2n\sum_{i=1}^{k}|g(c_i)-f(c_i)|\left(\frac{\epsilon}{4n\sum_{i=1}^{k}|g(c_i)-f(c_i)|}\right)\\ &<\frac{\epsilon}{2} \end{align*}

Is my proof of this theorem correct?

Nothing
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  • I see one technical issue, you've selected $\delta$ only to tailor to the needs of $|R(g,P)-R(f,P)|<\epsilon/2$ but actually you need $\delta$ small enough for both that and for $|R(f,P)-I(f)|<\epsilon/2$. Thus you need to choose $\delta_1,\delta_2$ for both those requirements and then take the minimum of the two $\delta$'s. Additionally, $\delta$ shouldn't directly depend on the number of points in the partition, i.e. $n$. (It will depend on the number of points where $g \neq f$, i.e. $k$.) – Ian Feb 26 '19 at 15:13
  • I get your first point. But for the second point, the reason i put $n$ inside $\delta$ because there is cases (maximally) that the points lies in two subintervals. (So $n$ points correspond to maximally 2n subintervals,) If i didn't put $n$ in $\delta$, I can't cancel off the $n$ – Nothing Feb 26 '19 at 16:11
  • You can avoid that problem more easily by scaling everything against the maximum of $|g(c_i)-f(c_i)|$ rather than the sum. Then it doesn't matter if two of the $c_i$ are somehow so close together that two of them are in the same subinterval, because that subinterval still contributes at most $\frac{\epsilon}{2k}$ to the sum over "bad" subintervals. There might then be fewer than $k$ "bad" subintervals, but that's not a problem. The way you did it now is actually circular, because actually you should be given $\delta$ and then $n$ is determined from the partition itself. – Ian Feb 26 '19 at 16:26
  • To avoid the clutter, assume wlog that $f$ and $g$ differ at only one point, $c$ Then, for a given partition $P$ wlog $c\in P$ and arrange it so that $x_i<c<x_{i+2}$ satisfies $x_{i+2}-x_i<\epsilon$, and the proof will fall out. – Matematleta Feb 26 '19 at 16:42
  • @Matematleta You can't make assumptions like "wlog $c \in P$" without invoking the equivalence of Riemann and Darboux integration. Without that, the only thing you can do is limit the mesh size $| P |$. (Now to be sure, invoking the equivalence of Riemann and Darboux integration is a perfectly good way to do this problem.) – Ian Feb 26 '19 at 16:57
  • @Ian Fair enough. The equivalence is a straightforward calculation. And worth doing IMO. In any case, one can certainly still assume $f$ and $g$ differ at only one point. – Matematleta Feb 26 '19 at 17:07
  • Let us see if f differs from g by only one point. Then that point has the possibility that this point lies at $x_i$ in the subintervals $[x_{i-1},x_i]$ and $[x_i,x_{i+1}]$. Then when doing the Riemann sum , I have to do it for two subintervals. This is my idea when it is one point. Then , for finitely many points, now i have $k$ points for which these $k$ points have the possibility to lie in $2k$ subinterval. (one point 2 subintervals). – Nothing Feb 26 '19 at 17:12
  • The point is that the restriction involving $\delta$ should just be $| P |<\delta$ for some number, but your form also requires the partition to be a particular size, which isn't how the definitions work. But I think really the problem is in your actual calculation: you should've gotten the bound $|R(g,P)-R(f,P)| \leq \sum_{i=1}^k |g(c_i)-f(c_i)| \delta$, so you need $\delta<\frac{\epsilon}{\sum_{i=1}^k |g(c_i)-f(c_i)|}$, and that's all. Note that this might have multiple $c_i$ in the same subinterval, but that just makes the bound less tight, not false. – Ian Feb 26 '19 at 17:13
  • By splitting the interval $[a, b] $ into a finite number of subintervals such that each $c_i$ is an end point of one subinterval we can reduce the problem to case where $k=1$ and $c_1$ is an end point $a$ or $b$. Now the proof is far simpler to type/write. – Paramanand Singh Feb 27 '19 at 01:43
  • @ParamanandSingh the statement you said is actually part (a) of this question . What I am stuck is how to generalise it to finitely many points . Is there anyway I can write to generalise it after I proved Part (a)? – Nothing Feb 27 '19 at 02:22
  • Read my previous comment. You have to split the interval $[a, b] $ using the points $c_i$. As an example for $k=2$ split like $[a, c_1],[c_1,d],[d,c_2],[c_2,b]$ and $a<c_1<d<c_2<b$. Now each subinterval has only one troublesome point $c_i$ and that is an end point of the subinterval. – Paramanand Singh Feb 27 '19 at 05:31
  • The problem is now c_1 is contained in two subintervals , which are [a,c_1] and [c_1,d]. When you do the sum, you need to do two times , one at the first subinterval , one at another subintervals . ( That's why we have maximally <2(g(c)-f(c))/delta – Nothing Feb 28 '19 at 05:26

1 Answers1

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Your approach via Riemann sums can be simplified with lesser use of symbolism. Let $M$ be a positive upper bound for both $|f|, |g|$ on $[a, b] $. Since $f$ is integrable, say with integral $I$, then for every $\epsilon>0$ we have a $\delta'>0$ such that for any partition $P$ of $[a, b] $ with norm $||P||<\delta'$ we have $$|S(f, P) - I|<\frac{\epsilon} {2}$$ where $$P=\{x_0,x_1,x_2,\dots,x_n\}$$ and $$S(f, P) =\sum_{i=1}^{n}f(t_i)(x_i-x_{i-1}),\, t_i\in[x_{i-1},x_i]$$ Since $g(x) =f(x) $ except for a finite number $k$ of exceptional points we have $$|S(g, P) - S(f, P) |<2Mk||P||$$ and thus if we choose $\delta$ such that $0<\delta <\min(\delta',\epsilon/(4Mk))$ we have $$|S(g, P) - S(f, P) |<\frac{\epsilon} {2}$$ whenever $||P||<\delta$ and thus $$|S(g, P) - I|<\epsilon$$ whenever $||P||<\delta$. This means that $g$ is also Riemann integrable on $[a, b] $ with the same integral $I$.

In short, changing the values of a function at a finite number of points affects neither its Riemann integrability nor the value of its integral (if it exists).

The word finite in above statement can't be replaced by infinite.

Paramanand Singh
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