# Pointwise convergence of probability densities imply weak convergence of probability measures

Assume $$P_n, n\in\mathbb{N}$$ and $$P$$ are absolutely continuous probability measures with respect to a sigma finite measure $$\mu$$ on $$(\mathbb{R},\mathcal{B})$$. Let $$f_n, n\in \mathbb{N}$$ and $$f$$ be the densities of above measures respectively.

I want to prove that if $$f_n$$ converges pointwise to $$f$$, then $$P_n$$ converges weakly to $$P$$. Here is what I have tried so far. I've read that even strong (pointwise) convergence follows, so I tried to prove that and the weak convergence follows:

Let $$A \in \mathcal{B}$$ and since $$f_n$$ are the the densities of $$P_n$$the following equalities hold $$P_n(A) = \int\limits_{A} f_n d\mu = \int 1_{A} f_n d\mu.$$ Now since $$f_n$$ converges pointwise to $$f$$, also $$1_{A} f_n$$ converges pointwise to $$1_{A} f$$. Here is where I'm stuck because I want to use the theorem about dominated convergence. Hence, I have to find an integrable bound for $$1_{A} f_n$$. If the theorem about dominated convergence applies I get

$$\lim\limits_{n\rightarrow\infty}P_n(A)=\int \lim\limits_{n\rightarrow\infty} 1_{A} f_n d\mu = \int 1_{A} f d\mu=\int\limits_{A} f d\mu = P(A)$$ which completes the proof.

Can someone help me to find a bound for $$f_n$$? Is this approach correct or am I missing something?

• does the converse hold?
– jay
Commented Feb 27, 2023 at 12:29

You can apply the lemma of Scheffé.

Observe that: $$\int\left(f-f_{n}\right)d\mu=\int fd\mu-\int f_{n}d\mu=1-1=0$$ implying that: $$\int\left(f-f_{n}\right)^{+}d\mu=\int\left(f-f_{n}\right)^{-}d\mu$$ and consequently: $$\int\left|f-f_{n}\right|d\mu=2\int\left(f-f_{n}\right)^{+}d\mu$$

Then for every measurable set $$A$$ we have:$$\left|P\left(A\right)-P_n\left(A\right)\right|=\left|\int_{A}\left(f-f_{n}\right)d\mu\right|\leq\int\left|f-f_{n}\right|d\mu=2\int\left(f-f_{n}\right)^{+}d\mu$$

Applying dominated convergence theorem we find: $$\lim_{n\to\infty}\int\left(f-f_{n}\right)^{+}d\mu=0$$ and conclude that: $$\lim_{n\to\infty}P_n\left(A\right)=P\left(A\right)$$

• does the converse hold?
– jay
Commented Feb 27, 2023 at 12:29
• No. If $P_n=P$ for every $n$ then $P_n$ converges weakly to $P$. But distinct densities for $P$ wrt the Lebesguemeasure exist. If $f,g$ denote two distinct densities and we let $f_n=g$ then we do not have $f_n\to f$ pointwise. Commented Feb 27, 2023 at 13:17
• @jay I see now that I forgot to address to you. See my former comment. Commented Feb 27, 2023 at 16:22