Show that $\left\{ f_{n}\right\}$ converges pointwise on $E$ to a function $f$ that is finite a.e. on $E$.

Question

Let $\left\{ f_{n}\right\}$ be an increasing sequence of nonnegative measurable function on $E$. Suppose there is a number $M>0$ such that $\left|\int_{E}f_{n}\right|\leq M$ for all $n$. Show that $\left\{ f_{n}\right\}$ converges pointwise on $E$ to a function $f$ that is finite a.e. on $E$ and $\lim_{n\to\infty}\int_{E}f_{n}=\int_{E}f.$

I think I only have to show that $\{f_n\}$ converges pointwise to $f$, since I can use MCT for the second part. In order to show the convergence, I have to show that for each $\epsilon>0$, for each $x\in E$, there exist $N$ such that $|f_n(x)-f(x)|<\epsilon$ for all $n>N$. This is the part where I actually dont know how to show. So, how can I use the given in order to proceed?

Answer 1

The pointwise limit $f$ surely exists as a measurable function from $E$ to $[0,+\infty]$ as pointwise we have an increasing real sequence that converges to its supremum.

You only have to argue that $A=\{x: f(x)=+\infty\}$ has measure $0$. If you've done that MCT (e.g. as given here, will do the rest.

So suppose it has not, and deduce a contradiction with the existence of $M$, bounding the integrals of the $f_n$.

Basic idea: If $\mu(A) > 0$ then certainly $f_n(x) > \frac{2M}{\mu(A)}$ for $x \in A$ and $n$ large enough, and then

$$\int_E f_n \,d\mu \ge \int_A f_n \,d\mu \ge \mu(A) \cdot \frac{2M}{\mu(A)}=2M$$

which would give a contradiction. Minor issue: as convergence is pointwise we can achieve the desired lower bound for $f_n$ only for a subset of $A$ (but these subsets have measures converging to $\mu(A)$). So some extra argumentation for that is needed.

Answer 2

$f_n$ converges (by monotonicity) to an extended real valued measurable function $f$. Fatou's Lemma shows that $\int f \leq \lim \inf \int f_n \leq M <\infty$. This implies that $f$ is finite a.e. on $E$ and Monotone Convergence Theorem gives $\lim \int_E f_n =\int_E f$.