# Where is the dominated convergence theorem being used?

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?

• Hint: Since the left hand side of the inequality is proportional to the variance of a random variable, where in the definition of the variance might an integral be involved? – whuber May 26 '15 at 15:29
• Thanks! So $nVar(\bar{X_n}) = n E (\bar{X_n} - \mu)^2$ and the integral comes form here, but we have a function that dominates the whole integral ($\sum_{h = -\infty}^{\infty} |\gamma(h)|$) and not only the pdf. Looking at the wikipedia article for the dominated convergence theorem I am talking about the $g$ function. en.wikipedia.org/wiki/Dominated_convergence_theorem. – Monolite May 26 '15 at 16:06
• @whuber Is this a dominated converge for sums? – Monolite May 26 '15 at 20:46
• I think that's the intention. The sum is, after all, a Lebesgue integral with respect to a counting measure supported on the natural numbers. – whuber May 26 '15 at 21:47

Assuming absolute summability of the autocovariance function (i.e. $\sum_{h=-\infty}^{\infty}|\gamma(h)| < \infty$) \begin{align*} \lim_n n \text{Var}(\bar{X}_n) &= \lim_n n^{-1} \sum_i \sum_j \text{Cov}(X_i,X_j) \\ &= \lim_n n^{-1} \sum_i \sum_j \gamma(|i-j|) \\ &= \lim_n n^{-1} \sum_{h \in \mathbb{Z}} (n-|h|) \gamma(h) \\ &= \lim_n \sum_h \left( 1-\frac{|h|}{n}\right) \gamma(h) \\ &= \sum_h \lim_n \left( 1-\frac{|h|}{n}\right) \gamma(h) \tag{1} \\ &= \sum_h \gamma(h). \end{align*}
1. For all $n$, $\left|\left( 1-\frac{|h|}{n}\right) \gamma(h)\right| \le |\gamma(h)|$
2. And $|\gamma(h)|$ is "integrable" (because we're assuming absolute summability).