Solving a problem using of Chebyshev's Inequality Let $f \in L_{1}(\mu)$ and let $M \gt 0$  such that $$|\frac{1}{\mu(E)}\int_{E}f d\mu| \le M$$ for every $E \in S$ with $0 \lt \mu(E) \lt \infty$. Show that $|f(x)| \lt M$ for a.e $x(\mu)$.
Let $F=\{x \in X$ such that $ |f(x)| \gt M\}$. I need to show that $\mu(F)=0$. Suppose it is not. Then $$\int_{F}|f|d\mu \gt M\mu(F)$$, But then I am unable to get any contradiction. 
Thanks for the help!!
 A: You were almost there!  (I will assume we're working with Lebesgue measure on $\mathbb{R}$ or something similar..)
First let me show that we can move the absolute value sign in the given inequality to the inside.  
For a measurable set $E$, write $E= E^+ \sqcup E^-$, where $E^+$ is the set where $f$ is positive on $E$ and $E^-$ is the set where $f$ is negative on $E$.  Assume both $\mu E^+>0$ and $ \mu E^- > 0,$ for otherwise it's trivial.  Then what's given is $$\int_{E^+} |f| d \mu = \left|\int_{E^+} f d \mu \right| \leq M \mu E^+,$$ and $$\int_{E^-} |f| d \mu = \left| \int_{E^-} f d \mu \right| \leq M \mu E^-.$$  Adding these, we conclude that, for all measurable $E$ with $0< \mu(E) < \infty$ we have $$\frac{\int_E |f| d \mu}{\mu(E)} \leq M .$$
Now I pick up where you left off.  Let $F = \{x : |f(x)| > M\}.$  Suppose $\mu(F) > 0$.  Then Chebyshev followed by what's given implies $$\int_F |f| d \mu > M \mu(F) \geq \mu(F) \frac{\int_F |f| d \mu}{\mu(F)},$$ which implies $1 > 1$, a contradiction!
(Important detail: the first inequality is indeed strict by positive definiteness almost everywhere of the integral.)
A: If $\mu \{f>M\} >0$ then $\frac 1 {\mu \{f>M\}} \int_{(f>M)}f d\mu >M$ because $\frac 1 {\mu \{f>M\}} \int_{(f>M)}(f-M) d\mu >0$. This is a contradiction. Similar argument shows $\mu \{f<-M\}= 0$ which completes the proof. [ The proof would be more complicated if you were consiering complex valued functions but I believe your function is real valued.
