Prove that $g$ is integrable on $[a, b]$ and that $\int_{a}^{b} g=$ $\int_{a}^{b} f .$

Question

Let $f$ be integrable on $[a, b]$ and let $S=\left\{s_{1}, s_{2}, \ldots, s_{k}\right\}$ be a finite subset of $[a, b] .$ Suppose that $g$ is a bounded function on $[a, b]$ such that $g(x)=$ $f(x)$ for all $x \notin S$. Prove that $g$ is integrable on $[a, b]$ and that $\int_{a}^{b} g=$ $\int_{a}^{b} f .$

my progress:

$$ I stucked on \text { Letting } h=g-f \text { . Then } h(x)=0 \forall x \notin S \text { . } $$

I don't really know any measure theory, but I am very confident that because this finite subset (I presume) will have measure zero, it does not impact the integral — FShrike, Jul 24 '21 at 20:53
Since you only want to show it for finite sets, a strategy you can take is to show it for the case when $S$ has exactly one element, and then use induction. — Mor A., Jul 24 '21 at 20:54
With a bit of cleverness, you can see that you only need to prove this for $k = 1$. Once you do this, proceed by induction on $k$. The base case $k = 0$ is trivial, and the inductive step follows from the $k = 1$ case. — Mark Saving, Jul 24 '21 at 21:47
Does this answer your question? Showing that g is integrable and $\int^b_a{f}$ = $\int^b_a{g}$ — AKP2002, Jan 14 '22 at 13:54

No Signals · Answer 1 · 2021-07-25T13:02:56.423

Let be $\epsilon>0$ be sufficiently small such that for each $i=1,...k$ $$[s_i-\epsilon,s_i+\epsilon]\subset[a,b]\mbox{ and } [s_i-\epsilon,s_i+\epsilon]\cap[s_j-\epsilon,s_j+\epsilon]=\emptyset$$ Let P be a partition of $[a,b]$ such that each $[s_i−ϵ,s_i+ϵ]$ is a subinterval. Let $|h|<M.$ The lower and upper sum are respectively $\mbox{ }L(h,P)\geq-Mk2\epsilon$ and $U(h,P)\leq Mk2\epsilon$. Then $$ U(h,P)-L(h,P)\leq 4Mk\epsilon$$ Since $\epsilon$ is arbitrary it has been prooved that there exists Partitions $P$ such that upper and lower sum are far apart less than a given $\epsilon$, hence $h$ is integrable and so is $g$. Moreover $$ -2Mk\epsilon\leq\int_{a}^{b} h\leq 2Mk\epsilon$$ hence $$\int_{a}^{b} h=0$$

Your statement follows from linearity. This is a special case of a more general property of Riemannian integral. One would define a set of mesure $0$ and would show that any integrable function over a rectangle vanishing except on a set of measure $0$ has its integral equals $0$.(see Munkres-Analysis on Manifolds-Theorem 11.3)

score 0 · Answer 2 · edited Jul 24 '21 at 22:20

First lets prove that $g$ is integrable on $[a,b].$

$g$ is bounded because $f$ is bounded.

Assume, without loss of generality, that $a\leq s_1 < s_2 < \cdots <s_k \leq b$.

We will take a look on the first Interval which is: $[a,s_1]$ if $a<s_1$ or $[s_1,s_2]$ if $a=s_1$, so let assume that $a<s_1$.

Note that $f\in R\left[a,s_{1}\right]$ because $f\in R\left[a,b\right]$ and $[a,s_1]\subseteq[a,b]$. In addition, $\forall \,a<x<s_1\,\,\,g|_{[a,x]}=f|_{[a,x]} \text{ and } g\in R[a,x].$

Now, let $\epsilon>0$. Define $x=s_1-\frac{\epsilon}{2\omega(g,[a,s_1])}$, $\omega(g,[a,s_1])= \sup\limits_{j\in[a,s_1]} g(j)- \inf\limits_{j\in[a,s_1]} g(j).$

Because $g\in R[a,x]$,by Riemann integral definition, there is partition $\pi_{1}$ of $\left[a,x\right]$ such that: $\omega\left(g|_{\left[a,x\right]},\pi_{1}\right)<\frac{\epsilon}{2}$, $\pi_1=\{ a = x_0 < x_1 < \cdots < x_n = x \}, \omega (g|_{[a,x]}, \pi_1) = \sum_{i=1}^n \omega(g,[x_{i-1},x_i]) \cdot\Delta x_{i}.$

Now take partition $\pi=\pi_{1}\cup\left\{ s_{1}\right\} $.

You get:

$$\omega\left(g|_{\left[a,s_{1}\right]},\pi\right)=\omega\left(g|_{\left[a,x\right]},\pi_{1}\right)+\frac{\epsilon}{2\omega\left(g,\left[a,s_{1}\right]\right)}\omega\left(g,\left[x,s_{1}\right]\right)<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.$$

So by definition, $g\in R[a,s_1].$

In the exact same way you can show that $g\in R[s_i,s_{i+1}, \, g \in R[s_n,b]$. So $g\in R[a,b]$. Now The fact that $\int_a^b g=\int_a^b f$ is immediately from linearity of the integral.

Prove that $g$ is integrable on $[a, b]$ and that $\int_{a}^{b} g=$ $\int_{a}^{b} f .$

2 Answers2