Q:

Show $f(x)=\int_E x^tg(t)d\mu(t)$ is continuous when $\mu$ is a general measure

Define the function $f:[0,1] \to \mathbb{R}$ by
$$
 f(x)=\int_E x^tg(t)d\mu(t)
$$
where $E \subset \mathbb{R^+}$,
$\mu$ is a nonnegative measure on $\mathbb{R}$ 
and $g:\mathbb{R} \to \mathbb{R}$ is a $\mu$-integrable function, that is $\int |g|d\mu < \infty$.
Is $f$ a continuous function of $x$?
I would be tempt to use the relation between absolute continuity and the lebesgue integral
but as the measure is not Lebesgue, it's of no use.
Is it possible to show that $f$ is continuous?
Does this need any extra assumptions?

A:

Use the dominated convergence theorem noting that $|x^t g(t)| \leq |g(t)|$, when $x\in [0,1]$. Then if $x_n\to \hat{x}$, you will have $x_n^t \to \hat{x}^t$, hence $f(x_n) \to f(\hat{x})$. Since this is true for all such sequences, $f$ is continuous.

