Integral of an increasing function over subset of unit interval Let $E\subset [0,1]$ with $\mu(E)=t$ and $f$ an increasing function. Prove that $\int_0^t f\leq \int_E f$.
The claim seems to be pretty clear to me and I can imagine why this has to be true but I have a hard time to prove it. Any hints on that?
 A: Hints:
Let $A_1=[0,t] \cap E$,  $A_2 = [0,t] \setminus E$ and $A_3 = E \setminus [0,t]$.
Note that $\mu A_2 = \mu A_3$ and $f(x_2) \le f(t)$ for all $x_2 \in A_2$ and 
$f(t) \le f(x_3)$ for all $x_3 \in A_3$.
You need to show that $\int_{A_2} f \le \int _{A_3} f$.
A: This result would be obvious if $E$ was an interval, or a finite union of intervals. To extend this to general $E$, use the fact that any measurable set can be approximated by a finite union of intervals. Namely, for any $\delta>0$, there exists a set $F$ so that $F$ is a finite union of intervals and $\mu(E\oplus F)<\delta$, where $E\oplus F=(E\setminus F) \cup (F\setminus E)$ is the symmetric difference of $E$ and $F$.
You then have 
$$
\Big|\int_E f\,d\mu-\int_F fd\mu\Big|\le \int_{E\oplus F}|f|
$$
Now, if $f\in L_1$, then the above integral would become arbitrarily small as $\mu(E\oplus F)\to 0$. (The result I am using is $f\in L^1$ implies for all $\epsilon>0$ that there exists a $\delta$ so $\mu(A)<\delta$ implies $\int_A |f|\,d\mu<\epsilon$, which you should try to prove.) Since $|\int_E f\,\mu -\int_F f\,d\mu|$ is arbitrarily small, and $\int_F f\,d\mu \ge \int_0^t f\,d\mu,$ you can show that $\int_E f\,d\mu \ge \int_0^t f\,d\mu$. 
I leave it to you to figure out what to do $f\not \in L^1$. 
A: Hint
Use simple functions that approximate $f$ on $[0,1]$.
