Show: $f_n \to f, a.e. \Rightarrow F_n(x) := \int_{(-\infty, x]}{f_n d\lambda^1} \to F_0$

Question

Let $(f_n)_{n\in \Bbb N_0} \subset \mathcal M_+(\mathcal B^1)$, so that $\forall n \in \Bbb N_0$, $\int_{\Bbb R}{f_n d\lambda^1} = 1$. For $n \in \Bbb N_0$ we define: $$F_n : \Bbb R \to [0,1]; \:\:\: F_n(x) := \int_{(-\infty, x]}{f_n d\lambda^1}$$

Now i have to show that: $$\lim_{n\to \infty}f_n = f_0, \:\:\: \lambda^1-a.e. \Rightarrow \forall x\in \Bbb R \lim_{n\to \infty} F_n(x) = F_0(x)$$

I also have to show whether the reverse direction is also true. Any ideas or tips on how to show these? Thanks in advance!

Answer 1

See the following answer for a generalisation of the dominated convergence theorem: https://math.stackexchange.com/a/2053719/27978.

Then fix $x$ and note that with $\phi_n = f_n \cdot 1_{(-\infty,x]}$, we have $\phi_n(x) \to \phi_0(x)$ ae. $x$, $0 \le\phi_n \le f_n$, $f_n(x) \to f_0(x)$ ae. $x$ and $\int f_n \to \int f_0$ (in fact, the integrals are all equal).

Then we can conclude from the above that $F_n(x) = \int \phi_n \to \int \phi_0 = F_0(x)$. Since $x$ was arbitrary, we are finished.