1. ## Integrability

Using these 2 facts for integrable functions f, g
$1. \int_a^b (f+g)=\int_a^b f+\int_a^b g
$

$2. \int_a^b cf=c\int_a^b f$for a constant $c$

Prove that any integrable function f can be changed at a finite number of points in the interval [a,b] without changing its integrability or its integral.

Start of Proof
I know that I only have to prove it for changing 1 point, and an induction argument will take care of the rest.
But how do I go about doing this?

I started with the definition of integrability
$sup{ }L(f,P)=inf{ }U(f,Q)$, where $P$ and $Q$ are partitions of $[a,b]$

and so dealing with only the sup and a partition $Pa<x_1<b)" alt="Pa<x_1<b)" /> and $Qa<x_2<b)" alt="Qa<x_2<b)" />
I need to prove
$sup{ }f[a,x_1](x_1-a)+sup{ }f[x_1,b](b-x_1)=sup{ }f[a,x_2](x_2-a)+sup{ }f[x_2,b](b-x_2)
$

How do I use the given facts to solve this problem. Or do I need a new approach

2. Originally Posted by I-Think
Using these 2 facts for integrable functions f, g
$1. \int_a^b (f+g)=\int_a^b f+\int_a^b g
$

$2. \int_a^b cf=c\int_a^b f$for a constant $c$

Prove that any integrable function f can be changed at a finite number of points in the interval [a,b] without changing its integrability or its integral.

Start of Proof
I know that I only have to prove it for changing 1 point, and an induction argument will take care of the rest.
But how do I go about doing this?

I started with the definition of integrability
$sup{ }L(f,P)=inf{ }U(f,Q)$, where $P$ and $Q$ are partitions of $[a,b]$

and so dealing with only the sup and a partition $Pa<x_1<b)" alt="Pa<x_1<b)" /> and $Qa<x_2<b)" alt="Qa<x_2<b)" />
I need to prove
$sup{ }f[a,x_1](x_1-a)+sup{ }f[x_1,b](b-x_1)=sup{ }f[a,x_2](x_2-a)+sup{ }f[x_2,b](b-x_2)
$

How do I use the given facts to solve this problem. Or do I need a new approach
I would assume the intended method was as follows. I'll assume for ease that $\alpha$ is continuous. You can generalize
Lemma: Let $c\in[a,b]$ and $\mathbf{1}_c$ the indicator function on $\{c\}$. Then, $\mathbf{1}\in\mathcal{R}(\alpha)$ on $[a,b]$ and $\displaystyle \int_a^b\mathbf{1}_c\text{ }d\alpha(x)=0$.
Proof: Note that evidently $\displaystyle L(P,\mathbf{1}_c,\alpha)=0$ for every partition $P$ of $[a,b]$. Thus, it suffices to show that for every $\varepsilon>0$ there exists some partition $P$ of $[a,b]$ such that $U\left(P,\mathbf{1}_c,\alpha\right)<\varepsilon$. To do this choose $\displaystyle x_1,x_2$ such that $\alpha(x_2)-\alpha(x_1)<\varepsilon$ and $c\in(x_1,x_2)$. It's clear then that if $P=\{a,x_1,x_2,b\}$ then $U\left(P,\mathbf{1}_c,\alpha\right)<\varepsilon$ from where integrability follows. The integral then follows by recalling that $\displaystyle \int_a^b\mathbf{1}_c\;d\alpha(x)=\sup_{P\in\wp}L(P ,\mathbf{1}_c,\alpha)=0$. The conclusion follows. $\blacksquare$
From this note then that if $f\in\mathcal{R}(\alpha)$ on $[a,b]$ and $\tilde{f}$ is such that $f-\tilde{f}=\left(f(c)-\tilde{f}(c)\right)\mathbf{1}_c$ where $f(c)\ne\tilde{f}(c)$ then evidently $\tilde{f}$ is continuous since by our lemma we have that $\displaystyle \frac{1}{f(c)-\tilde{f}(c)}\left(f-\tilde{f}\right)$ is integrable and thus so is $\left(f(c)-\tilde{f}(c)\right)\cdot\frac{1}{f-\tilde{f}(c)}\left(f-\tilde{f}\right)f=\tilde{f}$ and
$\displaystyle \int_a^b\tilde{f}(x)\text{ }d\alpha(x)=(f(c)-\tilde{f}(c))\int_a^b\mathbf{1}_c\text{ }d\alpha(x)+\int_a^bf(x)\text{ }d\alpha(x)=\int_a^bf(x)\text{ }d\alpha(x)$