Theorem $\mathbf{A.2.11}$ (Dominated convergence). Let $f_n : X \to \mathbb R$ be a sequence of measurable functions and assume that there exists some integrable function $g : X \to \mathbb R$ such that $|f_n(x)| \leq |g(x)|$ for $\mu$-almost every $x$ in $X$. Assume moreover that the sequence $(f_n)_n$ converges at $\mu$-almost every point to some function $f : X \to \mathbb R$. Then $f$ is integrable and satisfies $$\lim_n \int f_n \, d\mu = \int f \, d\mu.$$
I wanted to know if in the hypothesis $|f_n(x)| \leq|g(x)|$ above, if I already know that each $f_n$ is integrable, besides convergent, the theorem remains valid? Without me having to find this $g$ integrable?
Best Answer
This is an excellent question. For the theorem to apply, you need the $f_n$'s to be uniformly dominated by an integrable function $g$. To see this, consider the sequence $$ f_n(x) := \frac{1}{n} \mathbf{1}_{[0,n]}(x). $$ Clearly, $f_n \in L^1(\mathbb{R})$ for each $n \in \mathbb{N}$. Moreover, $f_n(x) \to 0$ as $n \to \infty$ for each $x \in \mathbb{R}$. However, \begin{align*} \lim_{n \to \infty} \int_{\mathbb{R}} f_n\,\mathrm{d}m = \lim_{n \to \infty} \int_0^n \frac{1}{n}\,\mathrm{d}x = 1 \neq 0. \end{align*}
Nevertheless, you are not in too much trouble if you cannot find a dominating function. If your sequence of functions is uniformly bounded in $L^p(E)$ for $1 < p < \infty$ where $E$ has finite measure, then you can still take the limit inside the integral. Namely, the following theorem often helps to rectify the situation.
In a sense, one can do without a dominating function when the sequence is uniformly bounded in a "higher $L^p$-space" and the domain of integration has finite measure.