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!

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)
