Proving the continuity of upper and lower integral.

Question

In this problem, we will prove that if $f: [a,b] \rightarrow \mathbb{R}$ is bounded, then the functions $G(x) = \overline{\int_{a}^x}f(t) dt$ and $H(x) = \underline{\int_{a}^x}f(t) dt$ are continuous. The method is divided into two steps. I will provide the solution for a), but I would like feedback on the epsilon-delta proof in b).

a) Assume that $|f(x)| \leq M $ for all $x$ in $[a,b]$. Show that for all $x_{1}$, $x_{2}$ in $[a,b]$, $|G(x_{1}) - G(x_{2})| \leq M \cdot |x_{1} - x_{2}|$ and the same holds for the $H$ function.

b) Use the definition of continuity to show that $G$ and $H$ are continuous in $[a,b]$

For a) we observe that (using the definition of upper and lower integrals)...

$$\begin{align*} % Begin an align* environment for multi-line equations |G(x_{1}) - G(x_{2})| &= \left| \overline{\int_a^{x_1}} f(t) dt - \overline{\int_a^{x_2}}f(t) dt \right| &= \left| \overline{\int_{x_1}^{x_2}}f(t) dt \right| \\ &\le\overline{\int_{x_1}^{x_2}}|f(t)| dt \\ &\le\overline{\int_{x_1}^{x_2}}M dt \\ &= M |x_1 - x_2|. \end{align*} % End the align* $$ Therefore, $$|G(x_{1}) - G(x_{2})| \le M |x_1 - x_2|. $$

We can use the same argument for $H.$

$$ $$

Proof in b)

If G is contionius then for every $\epsilon > 0 $ we can choose a $\delta > 0$ such that $\|x - a|< \delta$ implies $|G(x) - G(a)| < \epsilon$. Here $x,a$ are an anolous $x_{1}$ and $x_{2}$ from a), chosen to be in the interval $[a,b]$ , with $x \le b$

It follows from a) that $$|G(x) - G(a)| <M |x_{1} - x_{2}| = M |x-a| < M\cdot\delta$$ This means that we can choose $\delta = \frac{1}{M} \cdot \epsilon $ And we end up with $$|G(x) - G(a)| < \epsilon$$

Are there any potential flaws with the proofs?

Answer 1

Your proof of part a) is fine. You could shorten step b) by just saying: "by step a), $G$ and $H$ are $M$-Lipschitz hence (uniformly) continuous".

Part b) is slightly problematic in four ways:

1. begining with "if blah then..." is clumsy if you want to prove blah. Better say "in order to prove blah, let us prove that...".
2. there is a conflict of notations with your new $a$, distinct from the lower bound of $[a,b].$ Better keep the letters $x_1,x_2$ instead of $a,x.$
3. (it is not a flaw but you should mention it), what you are proving is not continuity at $a$ (or rather at $x_1∈[a,b]$ as previously said) since your $δ$ does not depend on it, but uniform continuity on $[a,b],$ which is a stronger property.
4. (a detail of proof-writing), instead of "$\dots<Mδ$ This means that we can choose $δ=ϵ/M$ and we end up with...", better write: "Let us choose $δ=ϵ/M.$ Then, it follows from a) that whenever $|x_1-x_2|\le\delta,$ we have $|G(x_1)−G(x_2)≤Mδ=ϵ$". (Because you cannot claim $\dots<M\delta$ before choosing $\delta.$) (incidentally, it is more careful to write $\le$ than $<,$ in case $M=G=0$).