Exercise 2.23 in Folland's real analysis text

[psie]

TL;DR Summary

I have some elementary doubts about two functions specified in an exercise in Folland's book. This exercise is supposed to establish the Lebesgue criterion for Riemann integrability.

23. Given a bounded function $f:[a,b]\to\mathbb R$, let $$H(x)=\lim_{\delta\to0}\sup_{|y-x|\leq\delta}f(y),\quad h(x)=\lim_{\delta\to0}\inf_{|y-x|\leq\delta}f(y).$$Prove Theorem 2.28b by establishing the following lemmas:
a) $H(x)=h(x)$ iff $f$ is continuous at $x$.
b) In the notation of the proof of Theorem 2.28a, $H=G$ a.e. and $h=g$ a.e. Hence $H$ and $h$ are Lebesgue measurable, and $\int_{[a,b]}H\,dm=\overline{I}_a^b(f)$ and $\int_{[a,b]}h\,dm=\underline{I}_a^b(f)$.

I'm working the above exercise on the Lebesgue criterion for Riemann integrability in Folland's real analysis text, i.e. a function is Riemann integrable on $[a,b]$ iff it is continuous a.e. on $[a,b]$. I think I know a solution, but I'm more concerned about elementary things. Are $h,H$ both functions from $[a,b]$? How can I reason that their domains really is $[a,b]$? Are they well-defined, i.e. how can we know the limits always exist?

Attempt: Note that if $x\notin[a,b]$ one would have to consider taking the supremum or infimum of the empty set at some point, with values $-\infty$ or $\infty$ respectively. Thus if the domain of $h,H$ wasn't $[a,b]$, both functions would be infinite on a set of nonzero measure, hence not integrable. Let us fix $x\in [a,b]$. Let us show the functions are well-defined, that is, that the limits exist. But note that e.g. $\sup_{|y-x|\leq\delta}f(y)$ is an increasing function of $\delta$, i.e. $\sup_{|y-x|\leq\delta}f(y)$ decreases with decreasing $\delta$. Since $f$ is bounded, we have for any $\delta$ that $|\sup_{|y-x|\leq\delta}f(y)|\leq |\sup_{y\in[a,b]}f(y)|<\infty$, which shows that $\sup_{|y-x|\leq\delta}f(y)$ is bounded. Hence the limit as $\delta\to 0^+$ exists and this is just the infimum of the range of $\sup_{|y-x|\leq\delta}f(y)$.

 

[psie]

Possibly a simpler way to argue about the domain of $h,H$ would be to simply say that if the $\sup$ or $\inf$ equals $\pm\infty$, then the limit wouldn't exist.

EDIT: I'm confident that the domain of $h,H$ is $[a,b]$, see e.g. under the header "Functions from metric spaces" in this Wikipedia article. $x$ is supposed to be a limit point of $[a,b]$. Since the set of limit points of $[a,b]$ is the set itself, we have $h,H:[a,b]\to\mathbb R$.

 

[nuuskur]

We must consider the interval $[a,b]$ as our universe. Meaning for instance, when we calculate $\sup _{|t-a|\leqslant \delta} f(t)$, what is really meant here is $\sup \{ f(t) \mid a\leqslant t \leqslant a+\delta\}$. We don't go outside in the wild, where things are not defined. By assumption $f$ is bounded, so this finite supremum exists. As we decrease $\delta$, the supremum decreases. Since $f$ is bounded from below, the limit is calculated by the monotonicity principle.

Since we take limit as $\delta\to 0$, in principle it could be that delta is big enough to cover some bigger interval than $[a,b]$, but it does not matter because eventually $\delta$ will be small enough where things are well defined and finite.