# 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).
• Nice answer. For the second point, that $\lvert \gamma(h)\rvert$ is "integrable", you mean that the (right, left or other) Riemann sum of this converges to its Riemann integral, which must be bounded by hypothesis? In this case, some continuity conditions must be imposed to the function $\gamma$, right? – Celine Harumi Mar 30 at 20:57