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I am reading Vitali theorem here (statement page 4 and proof page 8)

I am interested in the following part :

Let $0<r<\infty$. Let $X_n$ and $X$ be $L^r(\Omega)$ random variables such that

  • $X_n\stackrel{P}{\to} X$ (convergence in probability)
  • $X_n\stackrel{L^r}{\to}X$ (convergence in $L^r(\Omega)$).

Prove that $E(|X_n|^r)\to E(|X|^r)$.

I'm not sure how to prove it. In particular it is not true that $a^r-b^r\leq |a-b|^r$ (even when $a,b>0$).

Thus this question is not a duplicate of this or this

Mike
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Laurent Claessens
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1 Answers1

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It seems that the assumption $f_n\xrightarrow{n\rightarrow\infty}f$ in measure is not needed. Here we prove that

If $f_n\xrightarrow{n\rightarrow\infty}f$ in $L_r(\mu)$, then $\lim_n\|f_n\|_r=\|f\|_r$.

  • For $r\geq1$ the conclusion follows from $|\|f\|_r-\|f_n\|_r|\leq\|f-f_n\|_r$ since $f\mapsto\|f\|_r=\Big(\int_X|f|^r\Big)^{1/r}$ is a norm on $L_p(\mu)$ (identification mod 0 is done).

  • For $0< r<1$ we use the fact that $d_r(f, g)=\int_X|f-g|^r\,d\mu$ is a (complete) translation invariant metric on $L_r(\mu)$. The second assumption is equivalent to $$d_r(f_n,f)=\int_X|f_n-f|^r\,d\mu\xrightarrow{n\rightarrow\infty}0$$ The triangle inequality yields $$d_r(f_n,0)\leq d_r(f_n,f)+d_r(f,0)\xrightarrow{n\rightarrow\infty}d_r(f,0)$$ whence we obtain $$\limsup_n\int_X|f_n|^r\,d\mu\leq \int_X|f|^r\,d\mu.$$ An application of Fatou's Lemma yields \begin{align} d_r(f,0)&=\int_X|f|^r\,d\mu\leq \liminf_n\int_X|f_n|^r\,d\mu\\ &=\liminf_n d_r(f_n,0)\leq\liminf_n(d_r(f_n,f)+d_r(f,0))\\ &=d_r(f,0) \end{align} Putting this together we have that $\lim_nd_r(f_n,0)$ exists and $$\int_X|f|^r\,d\mu=d_r(f,0)=\lim_nd_r(f_n,0)=\lim_n\int_X|f_n|^r\,d\mu$$

Perhaps of more interest is the converse statement, that is,

If $f_n$ converges to $f$ in probability and $\|f\|_r\xrightarrow{n\rightarrow\infty}\|f\|_r$, then $\|f-f_n\|_r\xrightarrow{n\rightarrow\infty}0$.

For $r\geq1$ the conclusion follows from a result due to Riesz, M.:

For $r\geq1$, if $f_n\xrightarrow{n\rightarrow\infty} f$ $\mu$-a.s. and $\|f\|_r\xrightarrow{n\rightarrow\infty}\|f\|_r$, then $\|f-f_n\|_r\xrightarrow{n\rightarrow\infty}0$.

A proof of this result is given in this MSE posting.

In the present case, it is enough that every subsequence of $(f_n)$ admits a subsequent $(f_{n_k})$ which converges to $f$ in $L_r(\mu)$. Observe that as $f_n\xrightarrow{n\rightarrow\infty}f$ in $\mu$-measure, any subsequence $f_{n'}$ of $f_n$ has a subsequence $f_{n_k}$ which converges to $f$ $\mu$-a.s. Hence, by Riesz theorem, $\|f_{n_k}-f\|_r\xrightarrow{k\rightarrow\infty}0$.

What is new, perhaps, is the case $0<p<1$. We prove here the following result:

For $0<r<1$, if $f_n\xrightarrow{n\rightarrow\infty} f$ $\mu$-a.s. and $\|f\|_r\xrightarrow{n\rightarrow\infty}\|f\|_r$, then $\|f-f_n\|_r\xrightarrow{n\rightarrow\infty}0$.

Proof: Recall the inequality $(a+b)^r\leq a^r+b^r$ for $a,b\geq0$ and $0<r<1$. Then $$|f-f_n|^r\leq (|f|+|f_n|)^r\leq |f|^r+|f_n|^r$$ that is $$ g_n=|f|^r+|f_n|^r - |f-f_n|^r\geq0$$ An application of Fatou's lemma yields $$ \int_X 2|f|^r\,d\mu\leq \liminf_n\int_X\big(|f|^r+|f_n|^r-|f-f_n|^r\big)\,d\mu $$ Consequently $$\limsup_n\int_X|f-f_n|^r\,d\mu\leq0\qquad\blacksquare$$

The conclusion follows as in the case $r\geq1$, that is, it is enough to show that every subsequence of $(f_n)$ admits a subsequence $(f_{n_k})$ which converges to $f$ in $L_r(\mu)$.

Mittens
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