$\sum_{m=1}^{\infty }\frac{m}{2^n} \mu (E_{n,m}) \uparrow \int f d\mu$ Let $f \geq 0$ and $E_{m,n}=\{x :m/2^n \leq f(x) < (m+1)/2^n \}$
I need to show that as $n \uparrow \infty$
$\sum_{m=1}^{\infty }\frac{m}{2^n} \mu (E_{n,m}) \uparrow \int f d\mu$
My attempt:
I know that by fixing $n$, we have that on the set 
$E_{n,m}$ that $m/2^n \leq f(x)$ If we consider m to to finite , then $F=\mathbb{1}_{E_{n,m}}m/2^n$ is a simple function such that 
$F \leq f \implies \int F d \mu \leq \int f d \mu \Leftrightarrow \sum_{m=1}^{K }\frac{m}{2^n} \mu(E_{m,n}) \leq \int f d \mu $
Now taking K to the limit we get that $\sum_{m=1}^{\infty }\frac{m}{2^n} \mu (E_{n,m}) \leq \int f d \mu$ . 
It is clear that for a fixed n the sets $E_{n,m}$ are disjoint 
First I need to show that the L.H.S is increasing in $n$ (I have already shown it is always less than equal to $\int f d \mu$. Then I also need to show that the lim  of the L.H.S is indeed $\int f d \mu$.
I cannot show that L.H.S is increasing in $n$ and that the limit is indeed the R.H.S
Can you give me any hints on how to proceed. Please note that I am not looking for any complete solution
Thank you 
 A: 
The integral is defined as the supremum of functions of bounded functions with finite support.

I would work directly from this definition to show the result.  If your function had been bounded and of finite support, then your above analysis is almost complete.  Each $f_n$ would be composed of a finite set of functions that as $n$ goes to infinity, obtaining the supremum in the definition.
However, your function is more general.  Here is how I would proceed.  In general, for $f$ to be integrable in the first place, it cannot have a) much mass off at infinity, b) much mass above a certain point.  In terms of integrals:
$$\int_\mathbb{R} f d\mu = \int_{[-N, N]^c} f \chi_{\{f(x) \leq M\}} d\mu + \int_\mathbb{R} f \chi_{\{f(x) > M\}} d\mu + \int_{[-N, N]} f \chi_{\{f(x) \leq M\}} d\mu$$
where $[-N, N]^c$ is the complement of the interval $[-N, N]$.
Hopefully that is suggestive enough that you can finish from there, and flesh out the details of the above.
A: The L.H.S. is increasing with $n$
Use the equality $$E_{m,n}=E_{2m,n+1} \biguplus E_{2m+1,n+1}$$ to get $$\sum_{m=1}^{\infty }\frac{m}{2^n} \mu (E_{n,m}) \le \sum_{m=1}^{\infty }\frac{m}{2^{n+1}} \mu (E_{n+1,m}).$$
The limit is indeed the R.H.S.
Prove that if $f$ is any simple function, you will have $\sum_{m=1}^{\infty }\frac{m}{2^n} \mu (E_{n,m}) \ge \int f d\mu$ for $n$ large enough.
A: Hint:
If $f_{n}:=\sum_{m=1}^{\infty}m2^{-n}1_{E_{m.n}}$ then $f_{n}\left(x\right)=\lfloor f\left(x\right)2^{n}\rfloor2^{-n}$ for
a fixed $x\in\mathbb{R}$.
In general $2\lfloor a\rfloor\leq\lfloor2a\rfloor$ for each $a\in\mathbb R$ so that: $$f_{n}\left(x\right)=\lfloor f\left(x\right)2^{n}\rfloor2^{-n}\leq\lfloor f\left(x\right)2^{n+1}\rfloor2^{-n-1}=f_{n+1}\left(x\right)$$
