$ g: [0,\infty[ \rightarrow R, a \rightarrow \int_{0}^{\infty} e^{-ax} h(x) dx$ is defined, continious.., if h is lebegue-measurable

Question

Show that for a borel-measurable-lebesgue-integrable function $h:[0,\infty] \rightarrow \mathbb R$ the function $ g: [0,\infty[ \rightarrow \mathbb R, a \rightarrow \int_{0}^{\infty} e^{-ax} h(x) dx$ is 1. properly defined, 2. continuous and 3. evaluate the integral for $a \rightarrow 0$. We are taking the lebesgue-measure on $\mathbb R$.

1. I think (but I would gladly take a reasoning for that) $e^{-ax} h(x)$ is measurable, meaning that I can look at $ \int_{0}^{\infty} |e^{-ax}h(x)|dx$, which is defined. Since $|e^{-ax}| \le 1$, we get $|e^{-ax}h(x)| \le |h(x)|$ and so we get $\int_{0}^{\infty} |e^{-ax}h(x)|dx \le \int_{0}^{\infty} |h(x)|dx < \infty$ ($g$ is integrable). Therefore $\int_{0}^{\infty} |e^{-ax}h(x)|dx < \infty$, resulting into the integrability of $g$, so that $g$ is finite and therefore defined.

Does is make sense?

2. I have no idea, how to show, that $g$ is continious. I need help there.

3. If I could take the limit into the integral, then the result should be the Integral over $h(x)$. But I don't know here either, how to reason that.