Proving expectation and variance of a function of a random variable tends to a fix point

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}

• 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. – kimchi lover May 21 at 14:21
• 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? – user3107695 May 21 at 15:15
• $\text{Var}_{Q(X)}[X] = \int_{\mathcal{X}} (X - \mathbb{E}_{Q(X)}(X))^2 dq(x) = 0$, 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. – sudeep5221 May 22 at 2:48