The Dominated convergence theorem says the following: Let $f_{i} \in L^{1}(X, \Sigma, \mu)$ such that
- $f_{i}$ converges pointwise $\mu$-a.e. and
- There exists $g \in L^{1}(X, \Sigma, \mu)$ such that for all $i \in \mathbb{N}$ $|f_{i}| \leq g$ $\mu$-a.e.
Then there exists a function $f \in L^{1}(X, \Sigma, \mu)$ so we have $f:= \lim_{i \to \infty}f_{i}$ a.e. and $\lim_{i \to \infty}\int_{X}|f_{i} -f|d\mu = 0$
But how do I specifically test for the almost-everywhere conditions? For example the following problem:
$\lim_{n \to \infty}\int_{\mathbb{R}}\frac{n\sin(\frac{x}{n})}{x(x^{2}+1)}dx$
where $x \neq 0$
The $f_{n}$ converges pointwise to $\frac{1}{1+x^{2}}$ and we can also use that as a dominating function, i.e. $g(x)$. The overall result ends up being $\pi$.
BUT can I just assume pointwise convergence and a dominating function means the almost-everywhere condition is satisfied? I assume not, in which case I don't know how to specifically show that is satisifed.
Yes, pointwise convergence implies convergence almost everywhere. Note that convergence almost everywhere just means that pointwise convergence is allowed to fail in a (measurable) set of measure zero. If this set of 'bad points' is empty, we don't have anything to worry about.
Btw, you may want to pay extra attention to $x=0$ in your example.