Where is the dominated convergence theorem being used?

Question

I am trying to fully understand the proof of a theorem, I only have a problem with the application of the dominated convergence theorem. For the sake of completeness I will upload the whole statement and proof:

I focus only on the second part, the proof states:

And If $ \sum_{h = -\infty}^{\infty} |\gamma(h)| < \infty$ then the dominated convergence theorem gives:

$$\lim_{n \rightarrow \infty} n Var(\bar{X_n}) = \lim_{n \rightarrow \infty} \sum_{|h| < n} \Big( 1 - \frac{|h|}{n} \Big) \gamma(h) = \sum_{h = -\infty}^{\infty} |\gamma(h)| $$

I understand the proof up until the dominated convergence theorem is used, do we not need a Lebesgue integral to use it? And what are we using it on?

Answer 1

\begin{align*} n \text{Var}(\bar{X}_n) &= n^{-1} \sum_{i=1}^n \sum_{j=1}^n \text{Cov}(X_i,X_j) \\ &= n^{-1} \sum_{i=1}^n \sum_{j=1}^n \gamma(i-j) \\ &= n^{-1} \sum_{h = -(n-1)}^{n-1} (n-|h|) \gamma(h) \\ &= \sum_{h = -(n-1)}^{n-1} \left( 1-\frac{|h|}{n}\right) \gamma(h) \\ &= \sum_{h \in \mathbb{Z}} f_n(h) \end{align*}

where $f_n(h) := \mathbb{I}(|h| < n)\left( 1-\frac{|h|}{n}\right) \gamma(h)$. Notice that

1. $f_n \le |\gamma|$ pointwise for any $n$, and
2. $|\gamma|$ is "integrable" because it's the same as absolute summability in this case (i.e. $\sum_{h=-\infty}^{\infty}|\gamma(h)| < \infty$).

Taking the limit as $n \to \infty$ on everything and applying DCT gives us

$$ \lim_n \sum_{h \in \mathbb{Z}} f_n(h) = \sum_{h \in \mathbb{Z}} \gamma(h) $$

because $f_n \to \gamma$ pointwise as $n \to \infty$.

Answer 2

One way to think about the dominated convergence theorem is that it is one of those theorems that give sufficient conditions for when you can move a limit inside an infinite sum or integral. In this answer, I will give a slight variation in the explanation in the other answer, to frame things in this way. This is really the same thing that the other answer is showing, but it is re-expressed here in a way that may be more familiar to readers who think of the dominated convergence theorem in the way I have described it here.

As shown in the other answer here, you can write the relevant quantity of interest as:

$$n \mathbb{V}(\bar{X}_n) = \sum_{h \in \mathbb{Z}} f_n(h) \quad \quad \quad f_n(h) \equiv \mathbb{I}(|h| < n) \bigg( 1 - \frac{|h|}{n} \bigg) \gamma(h).$$

We also have the limiting function:

$$f(h) \equiv \lim_{n \rightarrow \infty} f_n(h) = \gamma(h).$$

Now, if we are allowed to move the limit inside the infinite sum then we would have:

$$\begin{align} \lim_{n \rightarrow \infty} n \mathbb{V}(\bar{X}_n) &= \lim_{n \rightarrow \infty} \sum_{h \in \mathbb{Z}} f_n(h) \\[6pt] &= \sum_{h \in \mathbb{Z}} \lim_{n \rightarrow \infty} f_n(h) \\[6pt] &= \sum_{h \in \mathbb{Z}} f(h) \\[6pt] &= \sum_{h \in \mathbb{Z}} \gamma(h). \\[6pt] \end{align}$$

What allows us to move the limit inside the infinite sum here (which is essentially an implicit interchange of limits) is the discrete version of the dominated convergence theorem. This says that we can move the limit inside the sum if $|f_n(h)| \leqslant f(h)$ for all $h \in \mathbb{Z}$ and $\sum_{h \in \mathbb{Z}} f(h) < \infty$. The first of these conditions clearly holds in this case. The proof you are looking at is saying that if $\sum_{h \in \mathbb{Z}} |\gamma(h)| < \infty$ then the latter condition also holds and so we can then apply the dominated convergence theorem to get the require result.