Exercise 1.6.15 (Locally integrable functions II)

Question

For each $h>0$, let $E_h$ be a measurable subset of $B(0,h)$ with the property $m(E_h) > cm(B(0,h))$ for some $c>0$ independent of $h$. Show that if $f:\mathbb{R}^d	o\mathbb{C}$ is locally integrable, and $x$ is a Lebesgue point of $f$ then
$$\lim_{h	o 0} \frac{1}{m(E_h)} \int_{x+E_h} f \ \mathrm{d}m = f(x).$$
Conclude that Theorem 1.6.19 implies Theorem 1.6.12.

Elu Thingol · Accepted Answer

By translation invariance of Lebesgue measure, we have that $m(B(0, h)) = m(B(x, h))$ for all $x \in \mathbb{R}^d$ since $B(x, h) = x + B(0, h)$. Note that from $E_h \subset B(0, h)$ 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 obtain

$$0 = \lim_{h 	o 0} \frac{1}{m(B(x, h)} \int_{B(x, h)} |f - f(x)| \, \mathrm{d}m \geq \lim_{h 	o 0} \frac{c}{m(E_h)} \int_{B(x, h)} |f - f(x)| \, \mathrm{d}m.$$

Since $E_h + x \subset B(x, h)$ combined with the fact that our integrand is non-negative, we have by monotonicity of the domain:

$$0 \geq \lim_{h 	o 0} \frac{c}{m(E_h)} \int_{B(x, h)} |f - f(x)| \, \mathrm{d}m \geq \lim_{h 	o 0} \frac{c}{m(E_h)} \int_{x + E_h} |f - f(x)| \, \mathrm{d}m.$$

By non-negativity of the integrand, we obtain

$$\lim_{h 	o 0} \frac{c}{m(E_h)} \int_{x + E_h} |f - f(x)| \, \mathrm{d}m = 0 \Rightarrow \lim_{h 	o 0} \frac{1}{m(E_h)} \int_{x + E_h} |f - f(x)| \, \mathrm{d}m = 0,$$

meaning that

$$\lim_{h 	o 0} \left|\frac{1}{m(E_h)} \int_{x + E_h} f- f(x) \, \mathrm{d}m ight| \leq \lim_{h 	o 0} \frac{1}{m(E_h)} \int_{x + E_h} |f - f(x)| \, \mathrm{d}m = 0,$$

and thus

$$\lim_{h 	o 0} \left| \frac{1}{m(E_h)} \int_{x + E_h} f- f(x) \, \mathrm{d}m ight| = 0 \Rightarrow \lim_{h 	o 0} \frac{1}{m(E_h)} \int_{x + E_h} f- f(x) \, \mathrm{d}m = 0,$$

therefore by the linearity of the integral, we have that

$$\lim_{h 	o 0} \frac{1}{m(E_h)} \int_{x + E_h} f \, \mathrm{d}m = \lim_{h 	o 0} \frac{f(x)}{m(E_h)} \int_{x + E_h} 1\, \mathrm{d}m = \lim_{h 	o 0} \frac{f(x)}{m(E_h)} \int_{\mathbb{R}^d} \mathbb{1}_{x + E_h}(y) \, \mathrm{d}m$$

the right integral obviously equalling $m(x + E_h) = m(E_h)$ by translation invariance, and thus we have

$$\lim_{h 	o 0} \frac{1}{m(E_h)} \int_{x + E_h} f \, \mathrm{d}m = f(x)$$

as desired.

Exercise 1.6.15 (Locally integrable functions II)

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Exercise 1.6.15 (Locally integrable functions II)

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