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



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

  • $\begingroup$ Your writtens are very clear thanks $\endgroup$
    – user519955
    May 27 '18 at 8:19
  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 $\limsup_{n\to +\infty}a_n=\lim_{N\to +\infty}s_N$.

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

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

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

  • $\begingroup$ 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

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