Convergence of measure in probability I know that if a random measure $P_n$ on $\mathbb{R}$ satisfies
$$
P_n([a,b])\to P([a,b]),
$$
for all intervals, then $P_n\Rightarrow P$, so that 
$$
P_n f \to P f,
$$
for all uniformly continuous $f$. My question is about when 
$$
P_n([a,b])\overset{\mathbb{P}}\to P([a,b]).
$$
In that case, does it for example hold that 
$$
P_n f \overset{\mathbb{P}}\to  P f,
$$
for $f$ uniformly continuous? In my example, $P_n$ has bounded support. 
 A: In all generality, this is false. If $X_n$ converges in probability to $+\infty$, $\delta_{X_n}[a,b] \to 0$ in probability for every $a$ and $b$.
But if $f=1$, $P_n(f) = 1$ does not converge to $0(f)=0$. 
The statement holds if $P_n$ are probability distributions and all have the same bounded support (say [0,1]). Indeed, let $f$ be uniform continuous on $[0,1]$, supposed bounded by 1 (wlog). Then $f$ can be approximated by staircase functions : $||f - \sum_{i=1}^k a_i\mathbf 1_{[b_i,c_i]}||_\infty\leq \epsilon$. Then $$|P_nf - Pf| \leq P_n|f - \sum_{i=1}^k a_i\mathbf 1_{[b_i,c_i]}| + P|f - \sum_{i=1}^k a_i\mathbf 1_{[b_i,c_i]}| + |P_n(\sum_{i=1}^k a_i\mathbf 1_{[b_i,c_i]}) - P(\sum_{i=1}^k a_i\mathbf 1_{[b_i,c_i]})| \leq \epsilon + \epsilon + \sum a_i |P_n[b_i,c_i] - P[b_i,c_i]|$$
For $n$ high enough, for every $i$, |P_n[b_i,c_i] - P[b_i,c_i]|>\epsilon/k with probability $<\epsilon/k$. So (union bound), the probability that no $i$ is such that $|P_n[b_i,c_i] - P[b_i,c_i]|>\epsilon/k$ is $>1-\epsilon$.
So with probability $>1-\epsilon$, $|P_nf - Pf|\leq 3\epsilon$, proving convergence in probability.
Note that this is for one given function only. We haven't proven convergence in probability of the random measures.
