0

Copying from this question, let $(S, \Sigma, \mu)$ be a measure space. (Part of) Scheffé's lemma states:

Suppose $\{f_n\}_{n \in \mathbb{N}}, f \in \mathscr{L}^1 (S, \Sigma, \mu)$ and $\lim_{n \to \infty} f_n(s) = f(s)$ $\forall s \in S$ or a.e. in S. Then $$\lim_{n \to \infty} \int_S |f_n - f| \mathrm{d}\mu = 0 \iff \lim_{n \to \infty} \int_S |f_n| \mathrm{d}\mu = \int_S |f| \mathrm{d}\mu.$$

Integrating over $S$ seems restrictive, in the sense that if $f_n$ and $f$ are density functions, $S$ is the sample space and $\mu$ is the probability measure, then the lemma only involves definite integrals. Is Scheffé's lemma true if you replace $S$ with $A \in \Sigma$, so that CDFs become relevant?

Appreciate guidance on why or why not.

johnsmith
  • 343

1 Answers1

1

If $f\in L_1$, then for any $A\in \Sigma$ we have $|f\cdot\mathbb 1_A|\leq |f|$ and hence $f\cdot \mathbb 1_A\in L_1$.

Now simply apply the Lemma to the functions $f_n\cdot \mathbb 1_A,f\cdot\mathbb 1_A$ to obtain $\begin{align*}&\lim\int_A|f-f_n|d\mu=\lim\int_S|f\cdot 1_A-f_n\cdot 1_A|d\mu =0 \\ \Leftrightarrow &\lim \int_A|f_n|d\mu=\lim \int_S|f_n\cdot1_A|d\mu\rightarrow \int_S|f\cdot1_A|d\mu=\int_A|f|d\mu\end{align*}$

Small Deviation
  • 2,296
  • 1
  • 3
  • 22