If $m(E) = \infty$, then is $n(\chi_{E} \ast \chi_{[0, 1/n]}) \rightarrow \chi_{E}$ pointwise a.e.? Let $E$ be a Lebesgue measurable set. If $m(E) < \infty$, then the Lebesgue Differentiation Theorem shows that $n(\chi_{E} \ast \chi_{[0, 1/n]}) \rightarrow \chi_{E}$ pointwise almost everywhere as $n \rightarrow \infty$. Is this still true when $m(E) = \infty$?
I was thinking of writing $E = \bigcup_{m = 1}^{\infty}E_{m}$ where $E_{m} = E \cap [-m, m]$. Then $\chi_{E}(x) = \sum_{m = 1}^{\infty}\chi_{E \cap [-m, m]}(x)$ and $$n(\chi_{E} \ast \chi_{[0, 1/n]})(x) = \frac{1}{1/n}\int_{x - 1/n}^{x}\sum_{m = 1}^{\infty}\chi_{E \cap [-m, m]}(t)\, dt = \sum_{m = 1}^{\infty}\frac{1}{1/n}\int_{x - 1/n}^{x}\chi_{E \cap [-m, m]}(t)\, dt$$
where the last step is because of the Monotone Convergence Theorem. Since $m(E \cap [-m, m]) < \infty$, it would be nice if I could justify $$\lim_{n \rightarrow \infty}\sum_{m = 1}^{\infty}\frac{1}{1/n}\int_{x - 1/n}^{x}\chi_{E \cap [-m, m]}(t)\, dt = \sum_{m = 1}^{\infty}\lim_{n \rightarrow \infty}\frac{1}{1/n}\int_{x - 1/n}^{x}\chi_{E \cap [-m, m]}(t)\, dt$$ but I can't seem to be able to apply Dominated or Monotone Convergence.
 A: You do not have $\chi_E = \sum_{m=1}^\infty \chi_{E\cap [-m,m]}$, because the $[-m, m]$ are not pairwise disjoint.
Depending on the version of Lebesgue's differentiation theorem that you know, your claim is more or less obvious:
One version states that for $f \in L_\rm{loc}^1$ there is a null set $N \subset \Bbb{R}^n$, so that for $x \in \Bbb{R}^n\setminus N$, we have
$$
\frac{1}{\lambda(B_\varepsilon(x))} \int_{B_\varepsilon (x)} |f(y) - f(x)| \,dy \rightarrow 0.
$$
If you apply this to $f = \chi_E$ (which is locally integrable, because it is bounded), you can easily get your claim.
If you only know the above statement for $f \in L^1$, you can derive the above version, as the result is local in nature. Indeed, apply the "weaker" version to $f_m := f \cdot \chi_{[-m,m]^n}$ for every $m \in \Bbb{N}$ to conclude that there is a null-set $N_m \subset \Bbb{R}^n$ such that
$$
\frac{1}{\lambda(B_\varepsilon(x))} \int_{B_\varepsilon (x)} |f_m(y) - f_m(x)| \,dy \rightarrow 0
$$
holds for all $x \in \Bbb{R}^n \setminus N_m$.
Set $N := \bigcup_m N_m$ and note that for $x \in [-m,m]^n$, the value of the integral
$$
\frac{1}{\lambda(B_\varepsilon(x))} \int_{B_\varepsilon (x)} |f(y) - f(x)| \,dy
$$
only depends upon the values of $f$ in $[-(m+1), m+1]^n$ (at least for $0<\varepsilon < 1$), so that the above version is implied by the "weaker" version.
The same also holds if you only know/want to use the
$$
\frac{1}{\lambda(B_\varepsilon(x))} \int_{B_\varepsilon (x)} f(y)  \,dy \rightarrow f(x)
$$
version of the Lebesgue differentiation theorem.
