The MGF is defined as $\mathbb E[e^{tX}]$ - in case this is finite for $t$ in a neighborhood of $0$, we say it exists.
Can there be a situation where the MGF exists but is not continuous (in $t$)? Specifically since we differentiate it and then evaluate it at $0$, can there ever be a situation where it is discontinuous at $0$?
I assume not, but the question is how to prove this.
First, note that the set of all $t$ such that $M_X(t)$ is finite forms an interval $I$ and that $M_X$ is convex on $I$ (see the answers here).
Second, note that every function that is convex on an interval $I$ is continuous on the interior of $I$ (see the answers here).
So, in order for $M_X$ to be discontinuous at $0$, $0$ has to be an endpoint of $I$. In your wording, this cannot be the case if $M_X$ exists.
Some say that $M_X$ always exists even if $M_X(t) = \infty$ for all $t \ne 0$. Then, of course, $M_X$ is discontinuous at $0$ (for example, see the answer here).
Let $(\Omega, \mathcal{F},ℙ)$ be a measure space and $X:\Omega\to ℝ$ be a random variable.
By definition we have $$ g(t):=\mathbb E[e^{tX}]=\int_\Omega e^{tX} dℙ $$
Define $f_t(w):=e^{tX(w)}$. Assume that for some $\tau>0$ we have that $\int_\Omega f_\tau dℙ<\infty$ (i.e. the moment generating function exists in a single point).
Since we have $f_{t'}<f_t$ for all $t'<t$, the law of dominated convergence now tells us that for any $t'<\tau$ we have that $$ \lim_{t\to t'}g(t)=\lim_{t\to t'}\int_\Omega e^{tX} dℙ= \int_\Omega \lim_{t\to t'}e^{tX} dℙ = \int_\Omega e^{(\lim_{t\to t'}t)X} dℙ = g(\lim_{t\to t'} t) $$
, i.e. $g$ is continuous in $t'$ by the sequential continuity theorem.
So really, all we need is for $g(t)$ to exist in a single point $\tau$, from which it follows that $g$ is continuous for all $t<\tau$.
