Convergence in expectation of a discontinuous function

Question

Consider a random variable $X\in \mathbb{R}^d$. Let ${\theta_m}$ be a sequence of real numbers that converge to $\theta$. Let $f(x,y)$ be a function that is not continuous. To be specific, fix, $x=a$, $f_a(y):= f(a,y)$ is not a continuous function in $y$. However, $E(f(X,y))$ is a continuous function in $y$. Example: take $f(x,y) = I(x\leq y)$, with the given fact $P(X\leq y)$ is continuous in $y$. Now, my question is: is it possible to show, $$E|f(X,\theta_m)-f(X,\theta)|\xrightarrow{}0?$$

It feels like the above should be true, but unable to show it rigorously. Any hint or help will be greatly appreciated.

Initially, I thought it would be straight-forward by interchanging limit and expectation, by virtue of Fubini's theorem,

$$\text{lim}_{m\xrightarrow{}\infty} E|f(X, \theta_m)-f(X,\theta)| = E|\text{lim}_{m\xrightarrow{}\infty}f(X, \theta_m)-f(X,\theta)|$$

But, can't proceed from here since $f$ is discontinuous.

Answer 1

Without more assumptions, it's not true.

Take $d=1$ and let $$f(x,y) = \begin{cases} x, & \text{if } y=0 \\ 0, & \text{otherwise} \end{cases}$$ Let $X$ be any nontrivial random variable with mean zero, such as $X= \pm 1$ with probability $1/2$ each. Then $E[f(X, y)]=0$ for all $y$, which is certainly continuous in $y$. But for any $y \ne 0$ we have $E|f(X,0) - f(X,y)| = E|X| > 0$.