I know a theorem called Polya's theorem:
$X_n \rightarrow X$ in distribution as $n\rightarrow \infty$ is equivalent to $\sup_{x} | F_n(x) -F(x)| \rightarrow 0$ as $n \rightarrow \infty$, where $F_n, F$ are distribution functions of $X_n$ and $X$, respectively.
Do you know where I can find out the proof for this theorem? or do you have hints to prove it?
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$\begingroup$The supremum should be over all $x$, not over all $n$, it's pointless.
This statement fails in general without assumption that the limiting function is continuous. Counterexamples are quite obvious. Say, a sequence $X_n$ with CDF $F_n(x)=x^n1_{(0,1)}$ converges to $X=1$ in distribution, but $\sup_x | F_n(x) -F(x)|=1$.
The proof is straightforward. Since $F$ is continuous, for any $k\geq 1$ there exist points $-\infty=x_0<x_1<\ldots<x_k=+\infty$ such that $F(x_i)=i/k$. Then we can bound $|F_n(x)-F(x)|$ from above and from below separately on every interval $x_{i-1}<x\leq x_i$: we use monotonicity of cdf's here. $$ F_n(x)-F(x) \leq F_n(x_i) - F(x_{i-1}) =F_n(x_i) - F(x_i)+1/k $$and$$ F_n(x)-F(x) \geq F_n(x_{i-1}) - F(x_i) =F_n(x_{i-1}) - F(x_{i-1})-1/k. $$So $$ \sup_{x\in\mathbb R} |F_n(x)-F(x)| \leq \max_{i=0,1,\ldots,k} |F_n(x_i)-F(x_i)|+1/k. $$Next for fixed $k$ we can use pointwise convergence $F_n(x_i)\to F(x_i)$ for all $i=1,\ldots,k-1$ that implies right maximum tends to zero, and next use that $k$ can be taken as large as we wish.
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