10
$\begingroup$

1) Let $\{f_n\}$ be a sequence of nonnegative measurable functions of $\mathbb R$ that converges pointwise on $\mathbb R$ to $f$ integrable. Show that

$$\int_{\mathbb R} f = \lim_{n\to \infty}\int_{\mathbb R}f_n \Rightarrow \int_{E} f = \lim_{n\to \infty}\int_{E}f_n $$

for any measurable set $E$

I know that $\int_{\mathbb R} f = \int_{\mathbb R \setminus E} f + \int_{E} f$ and $\int_{\mathbb R \setminus E} f \le \liminf_{n\to {\infty}}\biggr(\int_{\mathbb R \setminus E} f_n \biggr)$ from Fatau's Lemma.

I couldn't obtain $\int_{E} f = \liminf_{n\to \infty}\int_{E}f_n = \limsup_{n\to \infty}\int_{E}f_n$ and I have seen that inequality below for obtaining it but I couldn't understand. Could someone explain me please?

$$\liminf_{n\to \infty}\int_{\mathbb R \setminus E}f_n = \int_{\mathbb R}f-\limsup_{n\to \infty}\int_{E}f_n$$

2) It has been written "since $\int_Ef_n \le \int_Ef$ (this inequality from monotonicity I have understood) thus

$$\limsup\int_Ef_n \le \int_Ef$$

in proof of The Monotone Convergence Theorem in Royden's Real Analysis. I couldn't see why that inequality obtains.

Thanks for any help

Regards

$\endgroup$
1
4
$\begingroup$

Clearly $|f_n - f| \le f_n + f$ so define $g_n = f_n + f - |f_n - f| \ge 0$. We have $g_n \xrightarrow{n\to\infty} 2f$ pointwise.

Fatou's lemma applied on $g_n$ gives

\begin{align} 2\int_{\mathbb{R}} f &\le \liminf_{n\to\infty} \int_\mathbb{R} g_n \\ &= \liminf_{n\to\infty} \int_\mathbb{R} f_n + \liminf_{n\to\infty} \int_\mathbb{R} f + \liminf_{n\to\infty} \left(-\int_\mathbb{R}|f_n - f|\right) \\ &= 2\int_{\mathbb{R}} f - \limsup_{n\to\infty} \int_\mathbb{R}|f_n - f| \end{align}

so $\limsup_{n\to\infty} \int_\mathbb{R}|f_n - f| = 0$, which implies $\int_\mathbb{R}|f_n - f| \xrightarrow{n\to\infty} 0$.

Now for any $E \subseteq \mathbb{R}$ measurable

$$\left|\int_E f_n - \int_E f\right| = \left|\int_E (f_n - f)\right| \le \int_E |f_n - f| \le \int_\mathbb{R} |f_n - f| \xrightarrow{n\to\infty} 0$$

so $\int_E f_n \xrightarrow{n\to\infty} \int_E f$.

$\endgroup$
1
  • $\begingroup$ Your writtens are very clear thanks $\endgroup$
    – user519955
    May 27 '18 at 8:19
4
$\begingroup$
  1. We can show a stronger result, namely, that $\int_{\mathbb R}\left\lvert f_n-f\right\rvert\to 0$, by applying Fatou's lemma to $g_n:= \left\lvert f_n-f\right\rvert-f+f_n$.

  2. If $\left(a_n\right)_{n\geqslant 1}$ is a sequence of real number such that $a_n\leqslant t$ for all $n$, then $\limsup_{n\to +\infty}a_n\leqslant t$, because for all $N$, $s_N\sup_{n\geqslant N}a_n\leqslant t$ and $\limsup_{n\to +\infty}a_n=\lim_{N\to +\infty}s_N$.

$\endgroup$
1
  • $\begingroup$ Thanks for answer $\endgroup$
    – user519955
    May 27 '18 at 8:20
4
$\begingroup$

Good question. Here is another possible approach.

You already get $\int_{\mathbb R} f = \int_{\mathbb R \setminus E} f + \int_{E} f$ and $\int_{\mathbb R \setminus E} f \le \liminf_{n\to {\infty}}\biggr(\int_{\mathbb R \setminus E} f_n \biggr)$. So we have

$$ \lim_{n\to {\infty}}\int_{\mathbb R} f_n =\int_{\mathbb R} f \leq\liminf_{n\to {\infty}}\biggr(\int_{\mathbb R \setminus E} f_n \biggr)+ \int_{E} f. $$

Notice that for real sequences $a_n$, $\liminf_{n\to \infty}a_n = -\limsup_{n\to \infty}(-a_n)$ (see the "property" section in wiki), and $\lim_{n\to {\infty}}\int_{\mathbb R} f_n = \limsup_{n\to \infty}\int_{\mathbb R} f_n$. Thus the inequality above can be rewritten as $$ \limsup_{n\to {\infty}}\int_{\mathbb R} f_n \leq-\limsup_{n\to {\infty}}\biggr(-\int_{\mathbb R \setminus E} f_n \biggr)+ \int_{E} f, $$ or equivalently, $$ \limsup_{n\to {\infty}}\int_{\mathbb R} f_n +\limsup_{n\to {\infty}}\biggr(-\int_{\mathbb R \setminus E} f_n \biggr)\leq \int_{E} f, $$

Next we will use another property for limit superior, which says that for two sequences $a_n$ and $b_n$, we have

$$ \limsup_{n\to \infty}(a_n+b_n)\leq \limsup_{n\to \infty}a_n + \limsup_{n\to \infty}b_n. $$

Therefore we have $$ \limsup_{n\to {\infty}}\biggr(\int_{\mathbb R} f_n-\int_{\mathbb R \setminus E} f_n \biggr)\leq \limsup_{n\to {\infty}}\int_{\mathbb R} f_n +\limsup_{n\to {\infty}}\biggr(-\int_{\mathbb R \setminus E} f_n \biggr)\leq \int_{E} f, $$

that is, $$\int_{E} f\geq \limsup_{n\to {\infty}}\int_{ E} f_n.$$

For the last step, by Fatou's lemma we have $\int_{E} f\leq \liminf_{n\to {\infty}}\int_{ E} f_n$, thus

$$ \lim_{n\to {\infty}}\int_{ E} f_n = \liminf_{n\to {\infty}}\int_{ E} f_n = \limsup_{n\to {\infty}}\int_{ E} f_n = \int_{E} f. $$

$\endgroup$
3
  • $\begingroup$ Thanks for answer. I have a little confusion about one point. While we’re using $\limsup (a_n+b_n) \le \limsup a_n + \limsup b_n$ inequality, how could we say if $\limsup (a_n + b_n) \le x$ $(x \in \mathbb R)$ then $\limsup a_n + \limsup b_n \le x$ ? Is it from supremum properties? $\endgroup$
    – user519955
    May 27 '18 at 8:26
  • $\begingroup$ That's generally not true. But in my proof the order is, we are saying if $\limsup a_n+\limsup b_n≤x$ then $\limsup(a_n+b_n)≤x$ $\endgroup$
    – Wanshan
    May 27 '18 at 10:26
  • $\begingroup$ I have just realized thanks :) $\endgroup$
    – user519955
    May 27 '18 at 10:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.