# Doubt on integration exercise from Spivak's Calculus on Manifolds.

I've been studying integration in $\mathbb{R}^n$ in Spivak's Calculus on Manifolds, and I've started with the following problem: "Let $f,g:A\subset\mathbb{R}^n\to\mathbb{R}$ where $A$ is a closed rectangle in $\mathbb{R}^n$ be such that $f$ is integrable and $g=f$ except at finitely many points of $A$. Show that $g$ is integrable and that $\int_A g = \int_A f$". In what follows if $f$ is a bounded function in the set $S$, $M_S(f)=\sup\{f(x) : x \in S\}$ and $m_S(f)=\inf\{f(x) : x\in S\}$.

Now, I don't really know how to start. My idea was the following: given $\epsilon > 0$ I must show that there's a partition $P$ of $A$ such that $U(g,P)-L(g,P)<\epsilon$, where $U$ and $L$ denote the upper and lower sums respectively. Now suppose that the points where the functions differ are $x_1,\dots,x_n$. I can pick a partition $P$ such that for each point $x_i$ where the functions differ there is one rectangle $S_i$ contaning $x_i$ and no other of such points.

If $\mathcal{S}_P$ denotes the set of all subrectangles of $A$ partitioned by $P$, then let $U\subset \mathcal{S}_P$ denote the set of those $S_i$. Therefore, the upper sum of $g$ can be written

$$U(g,P)=\sum_{S\in \mathcal{S}_P} M_S(g)v(S)=\sum_{S\in\mathcal{S}_P\setminus U}M_S(g)v(S)+\sum_{i=1}^n M_{S_i}(g)v(S_i)$$

Now, since in the subrectangles $S \in \mathcal{S}_P\setminus U$ we have $g=f$, this is the same as

$$U(g,P)=U(f,P)+\sum_{i=1}^n M_{S_i}(g)v(S_i)$$

Simillarly, the lower sum can be written in the same way. Now, this gives us

$$U(g,P)-L(g,P)= U(f,P)-L(f,P) + \sum_{i=1}^n (M_{S_i}(g)-m_{S_i}(g))v(S_i).$$

Now, since $f$ is integrable, I know that I can find a partition such that $U(f,P)-L(f,P)$ is small as I want. If the partition doesn't contain different subrectangles with each $x_i$ in just one of them, I can refine it so that it happens, and it'll still satisfy the bound for the term $U(f,P)-L(f,P)$.

Now I just don't know what to do with the other term. Is this a good way to do this proof? Can someone give me just a hint on how to continue? If this is not a good way, I just want a hint on how to start it in a better way.

Yes, you are doing it the right way, but you missed at one point. The line $$U(g,P)=U(f,P)+\sum_{i=1}^n M_{S_i}(g)v(S_i)$$ is in incorrect, because you should include the difference $\sum_{i=1}^n (M_{S_i}(g)-M_{S_i}(f))v(S_i)$, and an analogous difference for the lower sum. Next, try to make the rectangles $S_i$ small enough such that these differences can be as small as you want and the prove is done. Hope I made it clear.

• Thanks @MateusSampaio, that's right I should include those differences. But now I'm with a silly doubt: can that difference $M_{S_i}(g)-M_{S_i}(f)$ be really made as small as desired? I thought on the following: consider $f,g : [0,1]\to\mathbb{R}$ given by $f(x)=x$ and $g(x)=x$ at all points except $1/2$, where we set $g(1/2)=1000$. Now, I think that for this pair of functions it wouldn't be possible. Where I'm making a mistake? Thanks again for the aid! – user1620696 Oct 17 '13 at 2:20
• You're right, at the example. The difference that I was referring to was considering the volume $v(S_i)$ from the rectangle. Although you cannot make the difference from the sups small as you want, these are limited and so you can take the volumes small enough to make the product as small as you want. – Mateus Sampaio Oct 17 '13 at 2:30
• Now I understood, it was taking into account the volumes. I think now I've got it! Thanks again. – user1620696 Oct 17 '13 at 2:38

Perhaps a faster proof is to show that $f-g$ is integrable and $\int_A f-g=0$. To prove this, let's first pick an $\epsilon$. Since both $f$ and $g$ are bounded, so is $f-g$, and we can assume a bound $M$. Now note that $f-g=0$ except on a finite point set $\Gamma=\{x_1,\cdots,x_k\}$. Let $P$ be any partition which contains subrectangles $S_i$ with $v(S_i)=\epsilon/Mk$ and $x_i\in$ interior $S_i$. Then $$U(f,P)-L(f,P)=\sum_{S\ne S_i}\underbrace{(M(S)-m(S))}_{=\ 0} v(S) + \sum_{i=1}^{k} \underbrace{(M(S_i)-m(S_i))}_{\le\ M}\frac{\epsilon}{Mk}\le \epsilon$$

Since $\epsilon$ was arbitrary, we obtain the desired result.