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Let $f_n$ denote a sequence of PDFs, and $F_n$ denote the corresponding sequence of CDFs. Given $L^1$ convergence of the PDFs to some PDF $f$,

$$\int_\mathbb{R} |f_n(x) -f(x)| dx \rightarrow 0$$

does this imply $L^2$ convergence of the corresponding CDFs to the corresponding CDF $F$,

$$\int_\mathbb{R} \left(F_n(x) - F(x)\right)^2 dx \rightarrow 0$$

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First note that $0\leqslant F(x),F_n(x)\leqslant 1$ for all $n,x$ so $\sup_{n,x}|F_n(x)-F(x)|=2$. It follows that $$\sup_{n,x}(F_n(x)-F(x))^2\leqslant 2|F_n(x)-F(x)|.$$ From $L^1$ convergence of $f_n$, continuity of $F_n, F$, and nonnegativity of $f_n,f$, we have for each $t\in\mathbb R$ \begin{align} |F_n(t)-F(t)| &= \left| \int_{-\infty}^t f_n(x)\ \mathsf dx -\int_{-\infty}^t f(x)\ \mathsf dx\right|\\ &\leqslant \int_{-\infty}^t |f_n(x)-f(x)|\ \mathsf dx\\ &\leqslant \int_{\mathbb R} |f_n(x)-f(x)|\ \mathsf dx\stackrel{n\to\infty}\longrightarrow 0, \end{align} and hence $F_n$ converges in distribution to $F$. In fact, as $F$ and $F_n$ are bounded, continuous, and monotone, it follows from Pólya's extension to Dini's theorem (cf. *Problems and Theorems in Analysis, I, p. 270) that $F_n$ converges uniformly to $F$ on any compact subset of $\mathbb R$. Since $F$ is a CDF, we may extend it to a continuous function $\overline F$ on the extended real line $[-\infty,\infty]$ by $\overline F(-\infty)=0$, $\overline F(\infty)=1$ (and similarly for $F_n$, and therefore $F_n$ converges uniformly to $F$ on $[-\infty,\infty]$. Since $F$ and $F_n$ are uniformly continuous, given $\varepsilon>0$ we may choose $N$ so that $n\geqslant N$ implies $$\sup_{x\in\mathbb R}|F_n(x)-F(x)|<\varepsilon. $$ Then $$\int_{\mathbb R} (F_n(x)-F(x))^2\ \mathsf dx\leqslant 2 \int_{\mathbb R}|F_n(x)-F(x)|\ \mathsf dx\stackrel{n\to\infty}\longrightarrow 0$$ so that $F_n$ converges to $F$ in $L^2$.

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    $\begingroup$ The convergence of $|F_n-F|\to{0}$ is uniform; thus, the limit and the integral can be exchanged by the dominated convergence theorem (by definition of uniformity $|F_n-F|\leq{g_n}\to{0}$). $\endgroup$ – Vossler Mar 31 '16 at 14:24
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    $\begingroup$ Wow. It's been really long since I last saw such a complete, hardcore $\varepsilon-\delta$-style proof. Not sure it was necessary here, but +1 for the patience! $\endgroup$ – Vossler Mar 31 '16 at 18:02
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    $\begingroup$ @Math1000 thanks for your efforts! So convergence of the integral then follows because $[-\infty, \infty]$ is compact and $|F_n - F|$ converges uniformly on this closed interval? $\endgroup$ – user3825755 Apr 1 '16 at 13:58
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    $\begingroup$ @user3825755 Just a minor clarification in response to Math1000's earlier comment: "The interchange in order of limit operations is justified by uniform convergence, not by dominated convergence": the uniform convergence criterion for exchanging integral and limit is actually a consequence of the DCT. $\endgroup$ – Vossler Apr 1 '16 at 18:07
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    $\begingroup$ Sorry that this has taken a while to reply. Isn't the critical requirement for uniform convergence to imply the desired convergence that the measure of the space over which is integrated is finite? For "normal" compact subsets of $\mathbb{R}$, this is of course so, but the measure of $[-\infty,\infty]$ is no longer finite. e.g. see answers in this question $\endgroup$ – user3825755 Apr 6 '16 at 11:59

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