I'm reading some lecture notes and the author defines the following:
Let $\mu_{n},\mu$
be probability measures on $\left(\mathbb{R}^{k},\mathcal{B}\left(\mathbb{R}^{k}\right)\right)$
, we say $\mu_{n}$
converge weakly to $\mu$
if $\int fd\mu_{n}\longrightarrow\int fd\mu$
for all continuous compactly supported functions $f:\mathbb{R}^{k}\to\mathbb{R}$
.
An almost identical definition appears in many text books with the change of requiring the same thing for any continuous bounded function. I couldn't find any reference which showed that it actually does suffice to look only at compactly supported functions. Is this actually true?
Best Answer
So as i said in the comment a usefull notion here is what's called "tendue" in french, i don't know the equivalent english mathematical word for that but the english translation of the common french word "tendue" could be taut, tense or tight. I don't want to use the wrong english word so let's just use tendue for this post.
edit : as said in the comments the english word for tendue is "tight".
A sequences $(\mu_n)$ is said to be tendue if for every $\varepsilon >0$ there is a compact $K$ such that $\mu_n(K^C)<\varepsilon$ for all $n$.
If your sequence $(\mu_n)$ is tendue then the two definitions are equivalent, this is because in that case $C_c(\mathbb R)$ is dense in $C_b(\mathbb R)$ for the $L^1(\mathbb R, \mu_n)$-norm uniformly in $n$. Phrased like this it might seems a little obscure but try to show it as an exercice, it's not hard. So what we want to show now is that if (using your definition) $(\mu_n)$ converges weakly to the probability measure $\mu$ then $(\mu_n) $ is tendue. This will prove that the two properties are equivalent.
Now suppose that $(\mu_n)$ is not tendue, so there exists an $\eta>0$ such that for every compact $K$ one have $\lim \sup \mu_n(K^C)>\eta$. Since $\mu$ is a probability measure for every $\varepsilon>0$ there exists a compact set $K$ such that $\mu(K)>1-\varepsilon$ and $\mu(K^C)<\varepsilon$. Take $\varepsilon=\eta/2$, there is a continuous compactly supported function $f$ such that $0\leq f \leq 1$ and $\int fd\mu>1-\varepsilon>1-\eta/2$. But we also have $\liminf \int f d \mu_n<1-\eta$, which is in contradiction with the fact that $\int fd\mu_{n}\longrightarrow\int fd\mu$. This is absurd so $(\mu_n)$ must be tendue.
But the situation is not as nice as you could think : if you only assume that $(\mu_n)$ converge to some measure (not necessarily a probability measure) then the two definition are not equivalent. The (now deleted) example of nicomezi was a good illustration : take $\mu_n=\delta_n$, according to your definition $\delta_n$ converges weakly to $0$, but with the definition using $C_b$ functions $\delta_n$ doesn't converges. However, if you suppose that $(\mu_n)$ is tendue and converge weakly to some measure $\mu$ (not necessarily a probability measure) then $\mu$ is a probability measure and the two definitions are equivalent. So the good notion here is the notion of tendue sequences.