# Sequence of measurable functions and limit

1. May 23, 2010

### complexnumber

1. The problem statement, all variables and given/known data

Given a $$\sigma$$-algebra $$(X,\mathcal{A})$$, let $$f_n : X \to [-\infty,\infty]$$ be a sequence of measurable functions. Prove that
the set $$\{ x \in X | \lim f_n (x) \text{ exists} \}$$ is in
$$\mathcal{A}$$.

2. Relevant equations

Let $$(X,\mathcal{A})$$ be a $$\sigma$$-algebra and $$M_{+}(X) :=$$ set
of all functions $$f : X \to [0,\infty]$$ which are
$$\mathcal{A}$$-measurable. Let $$f_n \in M_+(X)$$, then if $$\displaystyle \lim_{n \to \infty} f_n(x) \exists$$ $$\forall x \in X$$, then $$\displaystyle f := \lim_{n \to \infty} f_n \in M_+(X)$$.

Any $$f : X \to [-\infty,\infty]$$ can be written $$f = f_+ - f_-$$
where $$f_+(x) := \sup \{ f(x), 0 \}$$, $$f_-(x) := - \inf \{ f(x), 0 \}$$. So $$f$$ is $$\mathcal{A}$$-measurable iff $$f_+, f_- \in M_+(X)$$.

3. The attempt at a solution

Should I use the above two facts to show that there exists a measurable $$\displaystyle f = \lim_{n \to \infty} f_n$$ and hence anything $$f^{-1}(E) \in \mathcal{A}, E \in \mathbb{R}$$?

2. Aug 10, 2010

### Eynstone

Here's a sketch of a proof.
1. Define f(x) =lim f_n(x) if f_n(x) converges ,
= 0 otherwise.
2. Using Lebesgue's dominated convergence theorem, show that f is measurable.
3. If a measurable function exists on a set S, its characteristic function ,too, is measurable.