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I am curious about the following question: Suppose that $f(x;\theta)$ is a bounded function of $x$, where the domain of $x$ is $[a,b]\subseteq \mathbb{R}$, and $\theta$ is viewed as a parameter. If $f(x;\theta)$ is continuous in $\theta$, then under what condition(s) can we guarantee that the function $$F(\theta)=\int_a^b f(x;\theta)\,\mathrm{d}x$$ is continuous in $\theta$ ? Can anyone give me some help? Thanks so much!

OnoL
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    When you say that $f(x; \theta)$ is continuous in $\theta$ do you mean for fixed $x$? Similar question for bounded in $x$. And why don't you require that $f$ is also continuous in $x$? – Qiaochu Yuan May 29 '13 at 01:22
  • It would seem likely to me that $f(x;\theta)$ is intended to be continuous in $\theta$ for fixed $x$ and bounded for fixed $\theta$. The boundedness condition ensures that the integral always exists. I'm not sure why the continuity condition is there. – dfeuer May 29 '13 at 02:50
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    @user65018: what is the domain of $f$? It's clearly $[a,b] \times \text{something}$, but what is that something? – dfeuer May 29 '13 at 02:52

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Since we are looking at continuity in $\theta$, we should instead interpret $f(x;\theta)$ as a family of continuous functions of $\theta$ parametrized by $x \in [a,b]$. Since

$$ |F(\theta + \delta) - F(\theta)| \le \int_a^b | f(x; \theta + \delta) - f(x; \theta)| \, dx ,$$

then a sufficient condition for $F(\theta)$ to be continuous is that the family $f(x; \theta)$ is equicontinuous, when viewed as parametrized by $x$.

Christopher A. Wong
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