I have come across an apparent contradiction while working with the setwise topology of probability measures. Can someone please point out the mistake to me?
Let $(X,\mathcal{A})$ be a standard Borel space such as $\Bbb{R}$. Let $(p_n)_{n=1}^\infty$ be a sequence of probability measure on $(X,\mathcal{A})$.
If we see the $p_n$ as elements of $[0,1]^\mathcal{A}$, equipped with the product (i.e. setwise) topology, then since $[0,1]^\mathcal{A}$ is compact, $(p_n)_{n=1}^\infty$ has a convergence subsequence in $[0,1]^\mathcal{A}$. Let's call this subsequence $q_i=p_{n_i}$.
As stated for example here (and in other places), the setwise limit of probability measures is a probability measure, and so the sequence $q_i$ converges to some probability measure $q$.
Suppose now that the original sequence $(p_n)$ was a sequence of Dirac measures, in the form $\delta_{x_n}$ where all the $x_n$ are distinct points of $X$. Then $q_i$ is also in a similar form $\delta_{y_i}$ where $y_i=x_{n_i}$ are distinct points of $X$. By the reasoning above (if it is correct), for every measurable set $A$ of $X$, the sequence $(\delta_{y_i}(A))$ converges.
Now take as $A$ the countable union $\{y_1\}\cup\{y_3\}\cup\{y_5\}\cup...$, for all odd numbers. We have that $$ (\delta_{y_i}(A)) = \begin{cases} 1 & i \mbox{ odd} ; \\ 0 & i \mbox{ even} , \end{cases} $$ and so the sequence $(\delta_{y_i}(A))$ does not converge.
Where is the mistake?
$[0,1]^{\mathcal A}$ is a compact Hausdorff space but it is not metrizable. So you can only say that $(p_n)$ has a convergent subnet, not necessarily a convergent subsequence.