Cfxtrjtrk

Simple question. When are we allowed to exchange limits and integrals? I'm talking about situations like
$$lim_{varepsilonto0^+} int_{-infty}^infty dk f(k,varepsilon) overset{?}{=} int_{-infty}^infty dklim_{varepsilonto0^+} f(k,varepsilon).$$
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 $lim_{ntoinfty}f_n = f$ pointwise almost everywhere. Then, $f$ is an integrable function and $int |f-f_n| to 0$. In particular, $lim_{ntoinfty}int f_n = int f$ (by the triangle inequality). This can be written as
$$ lim_{ntoinfty}int f_n = int lim_{ntoinfty} f_n.$$

(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
$$lim_{epsilonto0^+}f(epsilon) = L iff lim_{ntoinfty}f(a_n) = Lquad text{for $mathbf{all}$ sequences $a_nto 0^+$.}$$

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 $lim_{epsilonto0^+}f_epsilon=f$ pointwise almost everywhere. Then, $f$ is an integrable function and $lim_{epsilonto 0^+}int f_epsilon = int f$. This can be written as
$$ lim_{epsilonto0^+}int f_epsilon = int lim_{epsilonto0^+} f_epsilon.$$

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

Let's look at it in a sample case. We want to prove by DCT that $$lim_{varepsilonto0^+} int_0^infty e^{-y/varepsilon},dy=0$$

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

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 $e^{-x}$.