# Moment generating function expansion proof

I am currently learning measure theoretic probability and I am trying to fill in the details for a sketch proof of the following well-known result:

Let $X$ be a real-valued random variable on the probability space $(\Omega, \Sigma, \mathbb{P})$. The moment generating function $M:\mathbb{R}\rightarrow \mathbb{R}\cup\{\infty\}$ is given by $$M(\theta)=\mathbb{E}(e^{\theta X}).$$ If $M(\theta)<\infty$ for all $\theta$ in some open neighborhood $(a,b)$ containing $0$, then for all $\theta\in(a,b)$, $$M(\theta)=\sum_{r=0}^\infty\dfrac{\mathbb{E}[X^r]}{r!}\theta^r.$$

The proof goes like this. First we Taylor expand $e^{\theta X}$ into $\sum_{r=0}^\infty\dfrac{(\theta X)^r}{r!}$. Then

$$M(\theta)=\int e^{\theta X} d\mathbb{P}=\int \sum_{r=0}^\infty\dfrac{(\theta X)^r}{r!}d\mathbb{P}= \sum_{r=0}^\infty \int\dfrac{(\theta X)^r}{r!}d\mathbb{P}=\sum_{r=0}^\infty\dfrac{\mathbb{E}[X^r]}{r!}\theta^r.$$

What I want to do is to justify the exchange of limit (i.e. the infinite sum) and integral above. The note simply says dominated convergence theorem. In order to use DCT though, I need to know that $\left|\sum_{r=0}^N \dfrac{(\theta X)^r}{r!}\right|<Y$ for all $N\in\mathbb N$ for some integrable function $Y$. However, I can't seem to think of such a function $Y$.

Apologies if this is supposed to be trivial.

Generally, the interval in which $M$ has a power series expansion (in powers of $\theta$) is symmetric about $0$ and is smaller than the set in which $M(\theta)<\infty$. Example: $X$ has the exponential distribution with parameter $\lambda>0$. In this case $M(\theta)={\lambda\over \lambda-\theta}$ for $\theta\in(-\infty,\lambda)$, but $M(\theta)=\sum_{n=0}^\infty {\theta^n\Bbb E[X^n]\over n!}$ only for $\theta\in(-\lambda,\lambda)$.