Question on Lebesgue point integration 
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!

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.
