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Given $f:\mathcal{X} \rightarrow \mathbb{R}$ is a continuous function and $\mathbb{E}_{Q(X)}[X] \rightarrow x^\star$ ($x^\star$ is a fix number), $\mathbb{V}\text{ar}_{Q(X)}[X] \rightarrow 0$. How can we prove that \begin{align} \mathbb{E}_{Q(X)}[f(X)] \rightarrow f(x^\star), \quad \text{and}\qquad \mathbb{V}\text{ar}_{Q(X)}[f(X)] \rightarrow 0? \end{align}

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  • $\begingroup$ You need some boundedness condition on $f$ for the desired result to hold. If, for instance, $EX^4=\infty$ and $f(x)=x^4$ you would be in trouble. $\endgroup$ – kimchi lover May 21 at 14:21
  • $\begingroup$ Hi, does it hold if I have the condition $f(x) < +\infty$? Actually my 𝑓 is a log of probability density function so I think it satisfy the boundedness condition you mentioned? $\endgroup$ – user3107695 May 21 at 15:15
  • $\begingroup$ $\text{Var}_{Q(X)}[X] = \int_{\mathcal{X}} (X - \mathbb{E}_{Q(X)}(X))^2 dq(x) = 0$, where $dq(x)$ is the probability measure of $Q(X)$. Since the integrand is always non-negative and so is the measure, it implies that the integrand is almost surely zero wrt the measure. This implies, $x = x^*$ almost surely under $Q(X)$. Since $f(x)$ is continuous, both the claims follow. $\endgroup$ – sudeep5221 May 22 at 2:48

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