Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and $K > 0$ such that $|f(x) -f(y)| \leq K \min\{1, |x-y|\}, x, y \in \mathbb{R}$. Also $X_n \xrightarrow[n \rightarrow \infty]{\text{a.s.}} X$. I am trying to prove that $f(X_n) \xrightarrow[n \rightarrow \infty]{L^1} f(X)$.
We have that $f$ is Lipschitz with constant $K$: $|f(x) -f(y)| \leq K |x-y|$ because $0\leq \min\{1, |x-y|\} \leq 1$. From this answer, I started with $$\mathbb{E}[|f(X_n) - f(X)|] \leq K \mathbb{E}[|X_n-X|]$$ but I don't know that the RHS goes to 0. Many thanks!
Your assumption on $f$ implies $$ \mathbb{E}[f(X_n)-f(X)] \le K \mathbb{E}[\min(|X_n-X|,1)] $$ The term in the expectation is bounded and converges a.s. to zero. By Lebesgue dominated convergence to expectation converges to zero, which proves to claim.