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: <math> \mathbb E f(X_n) \to \mathbb E f(X)</math> for every bounded continuous functional ''f''.
So it suffices to prove that <math> \mathbb E f(g(X_n)) \to \mathbb E f(g(X))</math> for every bounded continuous functional ''f''. For simplicity we assume ''g'' continuous. Note that <math> F = f \circ g</math> is itself a bounded continuous functional. And so the claim follows from the statement above. The general case is slightly more technical.
===Convergence in probability===
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