Let $h: \Bbb{R}\to \Bbb{R}$ be a Lebesgue measurable function with $\int |h| < \infty$ and define $$h^*: \Bbb{R}\to [0,\infty]: b \mapsto \sup_{t > 0} \frac{1}{2t}\int_{b-2t}^{b+2t}|h|$$
Show that $h^*$ is Borel-measurable.
Attempt: We show $\{x\in \Bbb{R}: h^*(x) > c \}$ is an open subset of $\Bbb{R}$ for all $c\in \Bbb{R}$.
If $h^*(b) > c$, then there is $\epsilon > 0$ such that $$\frac{1}{2\epsilon}\int_{b-\epsilon}^{b+\epsilon}|h| > c$$ Choose $\delta > 0$ such that $$\frac{1}{2(\epsilon+\delta)}\int_{b-\epsilon}^{b+\epsilon}|h| > c$$
If $|z-b|< \delta$, then $[z-(\epsilon+\delta), z+(\epsilon+\delta)] \supseteq [b-\epsilon, b+\epsilon]$ so $$h^*(z) \geq \frac{1}{2(\epsilon+\delta)}\int_{z-(\epsilon+\delta)}^{z+\epsilon+\delta}|h| \geq \frac{1}{2(\epsilon+\delta)}\int_{ b-\epsilon}^{b+\epsilon}|h| > c$$ so $b$ is an interior point of $\{h^* > c\}$, as desired.
Is the above correct?
From my pov your proof is ok and straight forward. Depending of what you already know $h^*$ is just a composition of measurable functions hence measurable itself. To see this consider the following measurable functions:
$$f(b) = \int_{a}^b |h| d\lambda$$ is absolutely continuous hence measurable for each $a \in \Bbb R$ by the general fundamental theorem of analysis.
For measurable $g_t$ the supremum $$\sup_{t > 0} g_t(b)$$ is measurable as well and with $$g_t(b) = \frac{f(b+2t) - f(b-2t)}{2t}$$ we have that all $g_t(b)$ are measurable being the composition of measurable functions and we get $$h^*(b) = \sup_{t>0}g_t(b)$$ so all together that $h^*$ is measurable.