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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?

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Is it an identity function multiplied by $\int_{E} g(t) d\mu(t)$? – Nimza Jun 19 '12 at 18:20
This is just $f(x) =c x$, where $c$ is the integral above (with $x=1$). – copper.hat Jun 19 '12 at 18:22
@Nimza Good eyes, I should have wrote $x^t$, not $x$. Thank you. – Nicolas Essis-Breton Jun 19 '12 at 18:22
@copper.hat I meant the integral of $x^t$, sorry. – Nicolas Essis-Breton Jun 19 '12 at 18:25
up vote 3 down vote accepted

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.

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What if every $t\in E$ is negative? – Byron Schmuland Jun 19 '12 at 18:34
@ByronSchmuland Good point. I forgot to mention that the $t$ are positive. Thank you. – Nicolas Essis-Breton Jun 19 '12 at 18:36

It's a consequence of the dominated convergence theorem. We just need to show sequential continuity. Let $x\in [0,1]$ and $\{x_n\}\subset [0,1]$ a sequence which converges to $x$. Let $g_n(t):=x_n^tg(t)$. Then $g_n$ is integrable, $g_n(t)\to x^tg(t)$ for all $t\in E$ and $|g_n(t)|\leq |g(t)|$, which is supposed to be integrable.

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Both your answer and the one of copper.hat are excellent. I throw a coin to decide the winner. – Nicolas Essis-Breton Jun 19 '12 at 18:35
Luck of the Irish... – copper.hat Jun 19 '12 at 18:45

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