Let $\mu$ be a measure on $((0,\infty),\mathbb{B}(0,1))$ so that $M(t):=\int_{(0,\infty)}e^{tx}\mu(dx)<\infty$ for $t\in (0,\epsilon]$. Show $\lim_{t\to 0}M(t)=\mu(0,\infty)$.
$f(x)=e^{tx}$ for $t\in(0,\epsilon]$ is continuous and therefore measurable. It is given that the integral is finite so $f(x)\in \mathscr{L}^1(\mu)$.
$f(t)=e^{tx}$ is continuous on $(0,\epsilon]$ for $x\in (0,\infty)$ (is this obvious or does it need to be shown?)
The problem is I cant find a dominating integrable function with $|f(x,t)|\leq w(x)\in \mathscr{L}^1(\mu)$.
If $w$ were to exist, would this be correct? I know $M$ would be continuous $$\lim _{t\to 0}M(t)=\lim_{t \to 0}\int_{(0,\infty)}e^{tx}\mu(dx)=\int_{(0,\infty)}\lim_{t\to 0}e^{tx}\mu(dx)=\int_{(0,\infty)}1\mu(dx)=\mu(0,\infty)$$
Actually for the one-sided limit $\lim_{t\rightarrow 0}M(t)$, it suffices to consider with arbitrary monotone decreasing sequence $(t_{n})$ such that $t_{n}\rightarrow 0$ and looking at $\lim_{n\rightarrow\infty}M(t_{n})$.
Then $\varphi_{n}(x)=e^{t_{n}x}$, $(\varphi_{n})$ is monotone decreasing and we can apply Monotone Convergence Theorem.