This question is regarding a special case of the Central Limit Theorem, the De Moivre-Laplace Theorem, which is introduced in Rick Durrett's Probability: Theorem and Examples, 5th edition, chapter 3. Let $X_1,X_2,...$ be i.i.d. with $\mathbb P(X_1 = 1) = \mathbb P(X_1 = -1) = 1/2$ and let $S_n = X_1 + ... + X_n$. He proves that $\mathbb P (a \sqrt{2n} \leq S_{2n} \leq b \sqrt{2n}) \to \int_a^b \frac{e^{x^2/2}}{\sqrt{2\pi}}dx$ as $m \to \infty$ using Stirling's formula, and I completely get his proof of this. However, he states in Theorem 3.1.3 that $\mathbb P (a \leq \frac{S_{m}}{\sqrt{m}} \leq b) \to \int_a^b \frac{e^{x^2/2}}{\sqrt{2\pi}}dx$, saying that "to remove the restriction to even integers, observe that $S_{2n+1} = S_{2n} \pm 1$. I do get the intuition that $S_{2m+1}/ \sqrt{2m+1}$ stays really close to $S_{2m}/ \sqrt{2m}$ as $m$ gets really large, but I am having some trouble trying to prove the thoerem rigorously in probability theoretic argument. We can write
$$\mathbb P (a \leq \frac{S_{2m}+1}{\sqrt{2m+1}} \leq b) = \frac{1}{2}\mathbb P (a \leq \frac{S_{2m}+1}{\sqrt{2m+1}} \leq b) + \frac{1}{2}\mathbb P (a \leq \frac{S_{2m}-1}{\sqrt{2m+1}} \leq b)$$
$$= \frac{1}{2}\mathbb P (a \leq \frac{S_{2m}}{\sqrt{2m}}\frac{\sqrt{2m}}{\sqrt{2m+1}} + \frac{1}{\sqrt{2m+1}} \leq b) + \frac{1}{2}\mathbb P (a \leq \frac{S_{2m}}{\sqrt{2m}}\frac{\sqrt{2m}}{\sqrt{2m+1}} - \frac{1}{\sqrt{2m+1}} \leq b).$$
However, I am not sure whether I can safely take the limit to the inside of $\mathbb P$ as I let $m \to \infty$. Could anyone tell me if I can take the limit to the inside, and if so, why? If I cannot, why can I not, and what is another method of solving this case for odd integers?
Thank you
Fix $a<b$. For every $\epsilon>0$, if $m$ is large enough, then
$$\mathbb P (a+\epsilon \leq \frac{S_{2m}}{\sqrt{2m }} \leq b-\epsilon ) \le \mathbb P (a \leq \frac{S_{2m+1} }{\sqrt{2m+1}} \leq b) \le \mathbb P (a-\epsilon \leq \frac{S_{2m}}{\sqrt{2m }} \leq b+\epsilon ) \,. $$ Now we can bound the liminf and limsup of the middle quantity and conclude using the continuity of the normal distribution function.