3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ does the converse hold? $\endgroup$
    – jay
    Commented Feb 27, 2023 at 12:29

1 Answer 1

4
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ does the converse hold? $\endgroup$
    – jay
    Commented Feb 27, 2023 at 12:29
  • 1
    $\begingroup$ 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. $\endgroup$
    – drhab
    Commented Feb 27, 2023 at 13:17
  • 1
    $\begingroup$ @jay I see now that I forgot to address to you. See my former comment. $\endgroup$
    – drhab
    Commented Feb 27, 2023 at 16:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .