1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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: $$\lim_{n\to\infty}P_n\left(A\right)\to P\left(A\right)$$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.