real analysis - 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.