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===Convergence in distribution===
We will need a particular statement from the [[portmanteau theorem]]: that convergence in distribution <math>X_n\xrightarrow{d}X</math> is equivalent to
: <math> \limsup_{n\to\infty}\Prmathbb E f(X_n) \in F)to \leqmathbb \PrE f(X\in F) \text{</math> for every closedbounded setcontinuous }functional F''f''.</math>
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''. Note that ''F = f \circ g'' is itself a bounded continuous functional. And so the claim follows from the statement above.
Fix an arbitrary closed set ''F''⊂''S′''. Denote by ''g''<sup>−1</sup>(''F'') the pre-image of ''F'' under the mapping ''g'': the set of all points ''x'' ∈ ''S'' such that ''g''(''x'')∈''F''. Consider a sequence {''x<sub>k</sub>''} such that ''g''(''x<sub>k</sub>'') ∈ ''F'' and ''x<sub>k</sub>'' → ''x''. Then this sequence lies in ''g''<sup>−1</sup>(''F''), and its limit point ''x'' belongs to the [[closure (topology)|closure]] of this set, <math display="inline"> X\in\overline{g^{-1}(F)} </math> <!-- <span style="text-decoration:overline">''g''<sup>−1</sup>(''F'')</span> --> (by definition of the closure). The point ''x'' may be either:
* a continuity point of ''g'', in which case ''g''(''x<sub>k</sub>'') → ''g''(''x''), and hence ''g''(''x'') ∈ ''F'' because ''F'' is a closed set, and therefore in this case ''x'' belongs to the pre-image of ''F'', or
* a discontinuity point of ''g'', so that ''x'' ∈ ''D<sub>g</sub>''.
Thus the following relationship holds:
: <math>
\overline{g^{-1}(F)} \ \subset\ g^{-1}(F) \cup D_g\ .
</math>
Consider the event {''g''(''X<sub>n</sub>'')∈''F''}. The probability of this event can be estimated as
: <math>
\Pr\big(g(X_n)\in F\big) = \Pr\big(X_n\in g^{-1}(F)\big) \leq \Pr\big(X_n\in \overline{g^{-1}(F)}\big),
</math>
and by the portmanteau theorem the [[limsup]] of the last expression is less than or equal to <math display="inline"> \Pr\left(X\in\overline{g^{-1}(F)} \right). </math> <!-- Pr(''X'' ∈ <span style="text-decoration:overline">''g''<sup>−1</sup>(''F'')</span>). --> Using the formula we derived in the previous paragraph, this can be written as
: <math>\begin{align}
& \Pr\big(X\in \overline{g^{-1}(F)}\big) \leq
\Pr\big(X\in g^{-1}(F)\cup D_g\big) \\
\le {} & \Pr\big(X \in g^{-1}(F)\big) + \Pr(X\in D_g) =
\Pr\big(g(X) \in F\big) + 0.
\end{align}</math>
On plugging this back into the original expression, it can be seen that
: <math>
\limsup_{n\to\infty} \Pr \big(g(X_n)\in F\big) \leq \Pr \big(g(X) \in F\big),
</math>
which, by the portmanteau theorem, implies that ''g''(''X<sub>n</sub>'') converges to ''g''(''X'') in distribution.
===Convergence in probability===
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