Is this theorem an extension of Scheffé Lemma

Question

Let $\nu_n(A)=\int_A f_n d\mu$ and $\nu(A)=\int_A f d\mu$ where $f_n, f$ are density functions. If $\nu_n(X)=\nu(X)$ and $f_n\to f$ ($\mu$- a.e) then $$\sup\limits_{A\in \mathcal A}|\nu_n(A)-\nu(A)|=\frac12 \int_X |f_n-f|d\mu \to 0, as\quad n\to \infty.$$

It looks like Scheffe' Lemma but I am not sure. Does anyone know this theorem and what textbook contains it? Or maybe give me some hints for the proof. Thanks https://en.wikipedia.org/wiki/Scheff%C3%A9%E2%80%99s_lemma

Answer 1

This is just rote application of definitions. The first equality comes from the definition of total variation by picking sets on which the two measures differ. You can verify it by proving $\leq$ and $\geq$. The convergence of the integral to 0 is dominated convergence. First observe that

$$\int_X |f_n-f|=\int_X [f_n-f]_+d\mu-\int [f_n-f]_-d\mu=\int_{X_+}(f_n-f)d\mu-\int_{X_-}(f-f_n)d\mu=\int_X f-\int_Xf_n=1-1=0.$$

Then since $[f-f_n]_+\leq f$, dominated convergence gives

$$\int_X |f_n-f|d\mu\leq 2\int_{X^+}[f_n-f]_+d\mu \rightarrow 0$$