Let $(f_{n})_{n}$ be a sequence of non-negative measurable functions. Suppose that $f_{n}$ decrease, in measure, in the sense: $$m(\{f_{n} > \epsilon\}) \leq \frac{1}{2^{n}},\quad \forall n \in \mathbb{N}, \forall \epsilon > 0.$$ Show that $m(\limsup_{n \to \infty}\{f_{n} > \epsilon\}) = 0$ and that $f_{n} \to 0$ a.e.
- a.e: almost everywhere.
I proved in a previous question that:
If $(E_{n})_{n}$ is a sequence of measurable sets, then $$ m(\liminf E_{n}) \leq \liminf m(E_{n}).$$
I thought of a way that would allow me to directly apply this property, but first I should prove that $f_ {n} \to 0$. Moreover, I don't know if that's a good idea. Can anybody help me?
First prove (or use as it is well known) that for an increasing sequence of measureable sets where $A_n \subset A_{n+1}$, we have $\lim_{n \to \infty}m(A_n) = m(\cup_{n=1}^\infty A_n)$.
This implies for a decreasing sequence of measurable sets where $A_{n+1} \subset A_n$ and at least one set has finite measure, we have $\lim_{n \to \infty}m(A_n) = m(\cap_{n=1}^\infty A_n).$
These properties are known as continuity of measure from below and above. Proofs are given here.
By definition,
$$\limsup_{n \to \infty} E_n = \bigcap_{n=1}^\infty \bigcup_{k \geqslant n} E_k$$
Taking $E_k = \{ f_k > \epsilon\}$ and $A_n = \cup_{k \geqslant n} E_k$, the sequence $(A_n)$ is decreasing and we have
$$m(\limsup_{n \to \infty} E_n) = m \left(\bigcap_{n=1}^\infty A_n \right) = \lim_{n \to \infty} m(A_n) \leqslant \lim_{n \to \infty} \left(\sum_{k=n}^\infty m(E_k) \right)$$
Now apply what you know about $m(E_k)$ and can infer about the convergence of the series $\sum m(E_k)$ to finish.