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!
 A: 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)
A: 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.
