I'm attempting to understand a proof of a weak version of the central limit theorem. I got through the proof, but it relied on a lemma that was left without a proof nor without any motivation. It is of course the lemma in the picture below. How does one prove this? Does anyone perhaps know where I may find a proof of it that is suited for someone who has had calculus and real analysis on an undergraduate level? It does not have to be super rigorous, as long as it's sufficient on some basic level.
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1$\begingroup$ This is not so much a lemma as the central fact: it is called a "continuity theorem," in this case for mgfs. (The corresponding result for characteristic functions is called Lévy's continuity theorem.) The proof is advanced and usually omitted from most undergraduate probability books. $\endgroup$– symplectomorphicCommented Jan 8, 2017 at 4:41
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$\begingroup$ Without knowing enough of the background to give technical details, this lemma shows that the sequence of moments converging to $M_Z$ implies that the corresponding sequence of distribution functions converges to the corresponding distribution function $F_Z$. It appears continuity is important to this, and I suspect the proof is equal parts topology and probability theory. $\endgroup$– Justin BenfieldCommented Jan 8, 2017 at 4:42
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$\begingroup$ @symplectomorphic I understand. The author should've omitted the entire proof of the central limit theorem instead. He essentially just provided an incomplete proof which does no good since he does not prove every step. I was primarily interested in the De Moivre–Laplace theorem anyway. Is that just a special case of the central limit theorem? $\endgroup$– DavidCommented Jan 8, 2017 at 5:05
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$\begingroup$ @David: yes, the De Moivre-Laplace theorem is a special case of the classical CLT, much easier to prove. Most authors omit the proof of the continuity theorem because it requires advanced analysis (the theory of Fourier and Laplace transforms). I think it's useful to see the CLT pop out of the mgf convergence + a Taylor series approximation, even if you don't have the tools to give a rigorous proof of the continuity theorem. $\endgroup$– symplectomorphicCommented Jan 8, 2017 at 5:09
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1$\begingroup$ In any case, this paper by Curtiss, "A note on the theory of moment-generating functions," gives a proof, but it is not self-contained; it appeals to other results in the literature. $\endgroup$– symplectomorphicCommented Jan 8, 2017 at 5:11
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