Integral and measure theory question Let E_n be a sequence of Lebesgue measureable sets in [0,1].  Suppose that for $0 \leq k \leq 1$ we have that 
$m(E_n \cap [0,r])= kr$
for any r, such that $0 \leq r \leq 1$. 
Prove that the 
$$\lim_{ n \rightarrow \infty} \int_{E_n} f(x) dx= k \int_{[0,1]} f(x) dx,$$ 
where $f \in L^1([0,1])$. 
I have attempted the following. 
$$k \int_{[0,1]} f(x) dx = \lim_{n \rightarrow \infty} \frac{m(E_n \cap [0,r])}{r} \int_{[0,1]} f(x)dx
= \lim_{n \rightarrow \infty} \int_{[0,r]} \frac{\chi_{E_n} (t)}{r} dt \int_{[0,1]} f(x) dx= \int_{[0,1]} \int_{[0,r]}\frac{\chi_{E_n} (t) f(x)}{r} dt dx.$$  
I want to somehow change the order of integration to change $\chi_{E_n}(t)$ to $\chi_{E_n}(x)$ (perhaps applying Fubini Tonelli), but I don't think its possible. 
*******Applying the comments suggestions********
Since step functions are dense in $L^1$ there exist $\phi_l \nearrow f$.  We note that we can apply DCT because $\phi_l \leq f$ and $f \in L^1$. 
$$\lim_{n \rightarrow \infty} \int_{[0,1]} \chi_{E_n}(x) \lim_{l \rightarrow \infty} \phi_l(x)dx=
\lim_{n \rightarrow \infty} \lim_{l \rightarrow \infty} \int_{[0,1]} \chi_{E_n}(x) \phi_l(x)dx.$$ 
I want the change the order of the limits to say that 
$$\lim_{n \rightarrow \infty} \lim_{l \rightarrow \infty} \int_{[0,1]} \chi_{E_n}(x) \phi_l(x)dx=
\lim_{l \rightarrow \infty} \lim_{n \rightarrow \infty} \int_{[0,1]} \chi_{E_n}(x) \phi_l(x)dx=
\lim_{l \rightarrow \infty} k \int_{[0,1]} \phi_l (x) dx = 
k \int_{[0,1]} f(x)dx.$$
I don't know how to justify that I can indeed change the order of the limits.  
 A: I do not see the importance of the index $n$ in this problem, you can show in fact that for every $n\in \mathbb{N}$, we have
$$\int_{E_n} f(x) dx = k \int_0^1 f(x) dx,$$
which implies the equality will also hold if we take the limit as $n\rightarrow \infty$. 
For now let us call $E_n$ simply $E$ since the equality does not depend on the index $n$. 
Define a measure $\mu_E:\mathcal{A} \rightarrow \mathbb{R}$ such that 
$$\mu_E (A) = m(A\cap E).$$ (check this is indeed a measure) 
From the assumption that 
$m(E\cap [0,r]) = kr,$ 
we can apply Dynkin $\pi - \lambda$ theorem to conclude that 
$$\mu_E (A) = m(A\cap E) = km(A)$$
for each $A\in \mathcal{A}$. Then the integral equality would hold trivially. 
Edit:  $m(E\cap [0,r]) = kr,$  implies that $$\mu_E (A) = m(A\cap E) = km(A)$$ for each $A\in  \mathcal{A_0}$, the collection of all finite union of intervals in $[0,1]$. If you are not familiar with the Dynkin $\pi - \lambda$ theorem, see here (I used it to show that two measures are the identical if they agree on each interval).
