If $X_i$ are random variables with $X_n\stackrel{L_p}{\to}X$ and $g$ is bounded and continuous, how to prove that $g(X_n)\stackrel{L_p}{\to}g(X)$?

Question

If $X_i$ are random variables with $X_n\stackrel{L_p}{\to}X$ and $g$ is bounded and continuous, how to prove that $g(X_n)\stackrel{L_p}{\to}g(X)$ ?

Answer 1

You can for example proceed as follows.

Convergence in $L_p$ implies convergence in probability. Since $g$ is continuous, it means that $g(X_n) \to g(X)$ in probability, and since function $x \to \|x\|^p$ is continuous, it means that $|g(X_n) -g(X)|^p \to 0$ in probability.

Moreover, family $\{|g(X_n)-g(X)|^p\}$ is uniformly integrable, because it's dominated by $(2M)^p$, where $M$ is bound of $|g|$, namely $M=\sup_{x \in \mathbb R}|g(x)| < \infty$ (we're using boundedness here)

Now, we have characterisation:

$ Y_n \to Y$ in $L_1$ iff and only if $Y_n \to Y$ in probability and $\{Y_n\}$ is uniformly integrable.

Using this with $Y=0$ and $Y_n = |g(X_n)-g(X)|^p$ we get that $|g(X_n)-g(X)|^p$ converges in $L_1$ to $0$, which means that $g(X_n)$ converges in $L_p$ to $g(X)$.

Edit: Or as follows. Since convergence in $L_p$ is metrisable (because it's norm convergence) it's sufficient to show that from everysubsequence $g(X_{n_k})$ we can take sub-subsequence $g(X_{n_{k_m}})$ such that $g(X_{n_{k_m}}) \to g(X)$ in $L_p$.

So take any subsequence $g(X_{n_k})$. It converges in probablity to $g(X)$ (argument as above). Hence we can take sub-subsequence $g(X_{n_{k_m}})$ converging almost surely to $g(X)$. What's left is to apply dominated convergence theorem to sequence $|g(X_{n_{k_m}}) - g(X)|^p$, which is bounded by $(2M)^p$ and converges almost surely to $0$ (because we've chosen such subsequence)