Question on Lebesgue point integration

Question

If $f(x)$ is finite at $x$ and $\lim\limits_{h\to 0}\frac{1}{h}\int_x^{x+h} |f(t)-f(x)|dt = 0$ then $x$ is called a Lebesgue point of function $f$.

a) If $f$ is continuous at $x$ then $x$ is a Lebesgue point.

b) Give an example Lebesgue point is not a continuous point.

c) If $f$ is Lebesgue integrable on [a, b] then a.e points in [a, b] are Lebesgue points.

My attempt: c) For each $r \in Q$, if $E_r = \{lim_{h\to 0}\frac{1}{h} \int_x^{x+h} |f(t)-r|dt = |f(x)-r|\}$ then $\lambda (E_r) = b-a$. Set $\cap E_r$ is the answer to the question. I have no idea whether this is true or not. Can any one give me some hints for part a) and b) also? Thank you in advance!

Answer 1

For c), it appears that you have the right idea. Let $r \in \mathbb{Q}$ and $f \in L^1$. From Lebesgue's differentiation theorem applied to the function $|f(t) - r|$, we have that for ae $x \in [a,b]$ that

$\frac{1}{h} \int_x^{x+h} |f(t) - r| dt \to |f(x) - r|$ as $h \to 0$.

In particular, let $E_r$ be the set of exceptional points at which the above fails to hold for each $r$. So $|E_r| = 0$, by assumption. Set $E = \bigcup_r E_r$. Suppose that $x \notin E$. Since $f(x)$ is finite, choose $r_n \to f(x)$, with $r_n \in \mathbb{Q}$. Since $x \notin E$, $x \notin E_{r_n}$ for any choice of n. With $\epsilon > 0 $ fixed, choose $n$ sufficiently large that $|r_n - f(x)| < \epsilon$. Now choose $\delta_n > 0$ so that when $0 < h < \delta_n$,

$| \frac{1}{h} \int_x^{x+h} |f(t) - r_n| dt - |f(x) - r_n| | < \epsilon$.

So, $\frac{1}{h} \int_x^{x+h} |f(t) - f(x)| dt = \frac{1}{h} \int_x^{x+h} |f(t) - r_n + r_n - f(x)| dt $

$\le \frac{1}{h} \int_x^{x+h} |f(t) - r_n| dt + |r_n - f(x)| < 3\epsilon $.

So $x$ is a Lebesgue point of $f$.

Edit: For completeness, here's a solution to a)

let $f \in L^1$ be continuous at $x$. Let $\epsilon > 0$. Choose $\delta > 0$ so that $|f(x) - f(t)| < \epsilon$ whenever $|x - t| < \delta$. Take $0 < h < \delta$.

Then $\frac{1}{h} \int_x^{x+h} |f(t) - f(x)| dt \le \frac{1}{h} \int_x^{x+h} \epsilon dt$

where the last inequality follows since $t \in [x, x+h]$. This shows a)

Answer 2

This proof I learned from Cohn's book, and it relies on the fact that finite borel measures are a.e. differentiable.

WLOG $f$ is borel measurable, else replace $f$ with $f_1 = f$ a.e. that is borel measurable. Fix an interval $(a,b)$.For $r \in \mathbb{Q}$ define the measure $$ \mu_r(A) := \int_A |f(x)-r| 1_{(a,b)}dx $$ Then $\mu_r$ is a finite borel measure for all $r$.

Let $\lambda$ the Lebesgue measure and $B_{h,x} = (x-h,x+h)$. For $B_{h,x} \subseteq (a,b)$ we have $$ \frac{1}{\lambda(B_{h,x})}\int_{B_{h,x}}|f(x)-f(y)|dy \leq \frac{1}{\lambda(B_{h,x})}\int_{B_{h,x}}|f(x)-r|dy + \frac{1}{\lambda(B_{h,x})}\int_{B_{h,x}}|r-f(y)|dy $$ $$ =|f(x)-r|+ \frac{1}{\lambda(B_{h,x})}\int_{B_{h,x}}|r-f(y)|dy. $$ Take the limsup as $h \to 0^+$ and we get $$ \limsup_{h\to 0^+}\frac{1}{\lambda(B_{h,x})}\int_{B_{h,x}}|f(x)-f(y)|dy \leq |f(x)-r| + D\mu_r(x) = 2|f(x)-r| $$ a.e. But $\mathbb{Q}$ is dense in $\mathbb{R}$ so the right can be made arbitrarily small.