# Dominated convergence theorem - what sequence?

Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like $$\lim_{\varepsilon\to0^+} \int_{-\infty}^\infty dk f(k,\varepsilon) \overset{?}{=} \int_{-\infty}^\infty dk\lim_{\varepsilon\to0^+} f(k,\varepsilon).$$ Everyone refers to either dominated convergence theorem or monotone convergence theorem but I'm not sure if I understand how exactly one should go about applying it. Both theorems are about sequences and I don't see how that relates to integration in practice. Help a physicist out :)

The statement of the dominated convergence theorem (DCT) is as follows:

"Discrete" DCT. Suppose $$\{f_n\}_{n=1}^\infty$$ is a sequence of (measurable) functions such that $$|f_n| \le g$$ for some integrable function $$g$$ and all $$n$$, and $$\lim_{n\to\infty}f_n = f$$ pointwise almost everywhere. Then, $$f$$ is an integrable function and $$\int |f-f_n| \to 0$$. In particular, $$\lim_{n\to\infty}\int f_n = \int f$$ (by the triangle inequality). This can be written as $$\lim_{n\to\infty}\int f_n = \int \lim_{n\to\infty} f_n.$$

(The statement and conclusion of the monotone convergence theorem are similar, but it has a somewhat different set of hypotheses.)

As you note, the statements of these theorems involve sequences of functions, i.e., a $$1$$-discrete-parameter family of functions $$\{f_n\}_{n=1}^\infty$$. To apply these theorems to a $$1$$-continuous-parameter family of functions, say $$\{f_\epsilon\}_{0<\epsilon<\epsilon_0}$$, one typically uses a characterization of limits involving a continuous parameter in terms of sequences:

Proposition. If $$f$$ is a function, then $$\lim_{\epsilon\to0^+}f(\epsilon) = L \iff \lim_{n\to\infty}f(a_n) = L\quad \text{for \mathbf{all} sequences a_n\to 0^+.}$$

With this characterization, we can formulate a version of the dominated convergence theorem involving continuous-parameter families of functions (note that I use quotations to title these versions of the DCT because these names are not standard as far as I know):

"Continuous" DCT. Suppose $$\{f_\epsilon\}_{0<\epsilon<\epsilon_0}$$ is a $$1$$-continuous-parameter family of (measurable) functions such that $$|f_\epsilon| \le g$$ for some integrable function $$g$$ and all $$0<\epsilon<\epsilon_0$$, and $$\lim_{\epsilon\to0^+}f_\epsilon=f$$ pointwise almost everywhere. Then, $$f$$ is an integrable function and $$\lim_{\epsilon\to 0^+}\int f_\epsilon = \int f$$. This can be written as $$\lim_{\epsilon\to0^+}\int f_\epsilon = \int \lim_{\epsilon\to0^+} f_\epsilon.$$

The way we use the continuous DCT in practice is by picking an arbitrary sequence $$\pmb{a_n\to 0^+}$$ and showing that the hypotheses of the "discrete" DCT are satisfied for this arbitrary sequence $$a_n$$, using only the assumption that $$a_n\to 0^+$$ and properties of the family $$\{f_\epsilon\}$$ that are known to us.

• Let's see if I understood this correctly, using the more specific problem I mentioned in the question. First, I find some integrable function $g$ s.t. $|f(k,\varepsilon)| \leq g(k), \forall k \in\mathbb{R}$ and all $\varepsilon$ between $0$ and some positive $\varepsilon_0$. Then I check if $f(k,\varepsilon) \to f(k,0)$ for all $k$ except perhaps on a set of measure zero. If it does, I can exchange the limit and the integral. If not, I can't. Did I get everything right? – PhysSE is Cancer Mar 30 at 23:45
• @IvanV.: Yes, that's correct! – Alex Ortiz Mar 31 at 0:23
• Alright, thank you, much appreciated! – PhysSE is Cancer Mar 31 at 2:00

Let's look at it in a sample case. We want to prove by DCT that $$\lim_{\varepsilon\to0^+} \int_0^\infty e^{-y/\varepsilon}\,dy=0$$

This is the case if and only if for all sequences $$\varepsilon_n\to 0^+$$ it holds $$\lim_{n\to\infty}\int_0^\infty e^{-y/\varepsilon_n}\,dy=0$$

And now you can use DCT on each of these sequences. Of course, the limiting function will always be the zero function and you may consider the dominating function $$e^{-x}$$.