Trouble with showing continuity of an integral. Let $f\in L^1(0,\infty)$. Define $g(t)=\int_0^\infty e^{-tx}f(x)dx$. Prove that $g$ is bounded and continuous on $[0,\infty)$ and $\lim_{t\to\infty}g(t)=0$.

$\lim_{t\to \infty} g(t) = \lim_{t\to \infty} \int_0^\infty e^{-tx}f(x)dx$. Can we say this is the same as $\int_0^\infty \lim_{t\to \infty} e^{-tx}f(x)dx = \int_0^\infty 0f(x)dx=0$?
Note that $\{e^{-tx}\}$ is a positive decreasing sequence so for all $x\in (0,\infty)$.  it obtains it maximum when $t=1$. So $|e^{-nx}f(x)| &lt; e^{-x}f(x)$ 
furthermore for each $t$ $e^{-tx}$ is a positive decreasing function so its supremum is on the interval $(0,\infty)$ is $1$ which implies  $|e^{-nx}f(x)| &lt; e^{-x}f(x)&lt;f(x) \in L^1$. In otherwords for $h_t=e^{-tx}f(x)$, $|h_t|\leq f(x)\in L^1$. Is $\{h_t\}\in L^1$? Yes, of course because it is bounded from above by a $L^1$ function. Thus we know that we CAN pass the limit through the integral and so $\lim_{t\to\infty}g(t)=0$.

Claim: $g(t)$ is bounded. Since $\{e^{-tx}\}\leq 1$ for all $t$ and all $x \in (0,\infty)$ we have $\int_0^\infty e^{-tx}f(x)dx\leq \int_0^\infty f(x)dx &lt;\infty$ since $f(x) \in L^1$.

Claim: $g(t)$ is continuous. I attempt to show this using an $\epsilon$, $\delta$ proof but run into problems.  

I wonder, how can I find $\delta$, is there another more convenient way to show that $g(t)$ is continuous, and lastly are there any other concerns with the validity of this proof? Thanks in advance!

This is what I have tried so far...

Let $\epsilon&gt;0$. We want to find a $\delta$ s.t. if $|t_2-t_1|&lt;\delta \implies |g(t_2)-g(t_1)|&lt;\epsilon$. Consider $|g(t_2)-g(t_1)|=|\int_0^\infty e^{-t_2x}f(x)dx-\int_0^\infty e^{-t_1x}f(x)dx|=|(e^{t_2}-e^{t_1})\int_0^\infty e^{-x}f(x)dx|\leq |(e^{t_2}-e^{t_1})||\int_0^\infty e^{-x}f(x)dx|$ and we want this quantity less than $\epsilon$. But we know that $e^{-x}f(x) \in L^1$ so we can let $|\int_0^\infty e^{-x}f(x)dx|=||h_n||_1=M$ some number in the reals. So we want $|e^{t_2}-e^{t_1}|&lt;\epsilon /M$.
$g(t)$ is continuous. This follows from the Dominated Convergence Theorem and the fact that $g(t_1)=\lim_{t\to t_1} g(t)= \int_0^\infty \lim_{t\to t_1} e^{-tx}f(x)dx= \int_0^\infty  e^{-t_1 x}f(x)dx$.
$g(t)$ is continuous. This follows from the Dominated Convergence Theorem and the fact that $g(t_1)=\lim_{t\to t_1} g(t)= \int_0^\infty \lim_{t\to t_1} e^{-tx}f(x)dx= \int_0^\infty  e^{-t_1 x}f(x)dx$.