Let $(X,\mathcal A,\mu)$ be a measure space and $\langle f_n\rangle_{n\in\mathbb N}$ a sequence of integrable functions that converges $\mu$-almost everywhere. By Lebesgue's Dominated Convergence Theorem, if $\langle|f_n|\rangle_{n\in\mathbb N}$ is bounded above by an integrable function, $\lim_{n\to\infty}f_n$ is integrable and
$$\int\!\lim_{n\to\infty}f_n\;\mathrm d\mu=\lim_{n\to\infty}\int\!f_n\;\mathrm d\mu\quad.$$
LDCT cannot be used if we do not assume the said boundedness. For instance, if $\mu$ is the Lebesgue measure and $f_n=nI_{[0,1/n]}$, we get $0=1$.
But is the equality false whenever we drop this hypothesis? In other words, can "if" be replaced by "if and only if" on the theorem statement? Or is there an integrable non-dominated $\mu$-a. e. convergent sequence (and a measure) for which the equality occurs?
Best Answer
The converse does not hold. Take $X = [0,\infty)$ with Lebesgue measure. Let $a_n$ be a sequence of positive numbers such that $a_n \to 0$ but $\sum_n a_n = \infty$. (For instance, $a_n = 1/n$ would work.) Let $b_n = \sum_{i=0}^n a_k$. Then let $f_n = 1_{[b_{n-1}, b_{n})}$. Then $f_n \to 0$ a.e. and $\int f_n dm = a_n \to 0$ also. But if $f$ is a dominating function, we must have $f \ge \sup_n f_n = 1$ which is not integrable.
For a necessary and sufficient condition for convergence, look up the Vitali convergence theorem.