Lebesgue Differentiation Theorem-Type Problem I just read over a proof of the Lebesgue differentiation theorem in $\mathbb{R}^n$ and attempted a related exercise in the book that I am using (the Tao Measure Theory book). I am unable to figure this out and was hoping to get some tips or hints:
For each $h > 0$, let $E_h \subset B(0, h)$ such that $m(E_h) \geq c \cdot m(B(0, h))$ for some $c > 0$ which is independent of $h$. If $f \colon \mathbb{R}^n \to \mathbb{C}$ is locally integrable, with $x$ a Lebesgue point of $f$, then prove that
$$\lim_{h \to 0} \frac{1}{m(E_h)} \int_{E_h + x} f(y) \, dy = f(x).$$
I know that this is something that follows from the Lebesgue differentiation theorem (LDT), but after a few steps I'm not really sure where to go. I started by noting that by translation invariance we have that $m(E_h) \geq c \cdot m(B(x, h))$ so since we have
$$\lim_{h \to 0} \frac{1}{m(B(x, h))} \int_{B(x, h)} f(y) \, dy = f(x),$$
by the LDT it follows that
$$\lim_{h \to 0} \frac{c}{m(E_h)} \int_{B(x, h)} f(y) \, dy \leq f(x),$$
but I have zero idea where to go from there. I'm not sure how to change the domain of integration without complicating things further, and I don't know how to get the other inequality either. Any tips or suggestions would be great!
 A: Wow, I feel stupid for not getting this sooner but never mind. By translation invariance of Lebesgue measure, we have that $m(B(0, h)) = m(B(x, h))$ for all $x \in \mathbb{R}^n$ since $B(x, h) = x + B(0, h)$. Note that if $E_h \subset B(0, h)$ then it follows that $x + E_h \subset B(x, h)$. Thus putting these two facts together along with the assumption that $m(E_h) \geq c \cdot m(B(0, x))$ for some $c > 0$, we see that $m(E_h) \geq c \cdot m(B(x, h))$ for all $x \in \mathbb{R}^n$. Then by the Lebesgue differentiation theorem, we have that that
$$\lim_{h \to 0} \frac{c}{m(E_h)} \int_{B(x, h)} |f(y) - f(x)| \, dy \leq \lim_{h \to 0} \frac{1}{m(B(x, h)} \int_{B(x, h)} |f(y) - f(x)| \, dy = 0$$
And since (E_h + x \subset B(x, h),$ since out integrand is nonnegative we have by monotonicity of domain that
$$\lim_{h \to 0} \frac{c}{m(E_h)} \int_{x + E_h} |f(y) - f(x)| \, dy \leq \lim_{h \to 0} \frac{c}{m(E_h)} \int_{B(x, h)} |f(y) - f(x)| \, dy \leq 0$$
And then obviously since our integrand is nonnegative this implies that
$$\lim_{h \to 0} \frac{c}{m(E_h)} \int_{x + E_h} |f(y) = f(x)| \, dy = 0 \Rightarrow \lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} |f(y) - f(x)| \, dy = 0,$$
meaning that
$$\left|\lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y)- f(x) \, dy \right| \leq \lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} |f(y) - f(x)| \, dy = 0,$$
and thus
$$\left|\lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y)- f(x) \, dy \right| = 0 \Rightarrow \lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y)- f(x) \, dy = 0,$$
therefore by linearity of the integral we have that
$$\lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y) \, dy = \lim_{h \to 0} \frac{f(x)}{m(E_h)} \int_{x + E_h} \, dy = \lim_{h \to 0} \frac{f(x)}{m(E_h)} \int_{\mathbb{R}^d} \mathbb{1}_{x + E_h}(y) \, dy$$
the right integral obviously equalling $m(x + E_h) = m(E_h)$ by translation invariance, and thus we have that
$$\lim_{h \to 0} \frac{1}{m(E_h)} \int_{x + E_h} f(y) \, dy = f(x).$$
