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Let $(f_n)^\infty_{n=1}$ be a sequence of Lebesgue integrable functions on $[0,1]$ such that $f_n$ converges to $f$ almost everywhere in $[0,1]$. Suppose further

(a). $\sup_n\int_0^1|f_n|d\mu<\infty$;

(b). for any $\epsilon>0$, there exists $\delta>0$ such that for any measurable subset $E$ of $[0,1]$ with $\mu(E)<\delta$, $\sup_n\int_E|f_n|d\mu<\epsilon$.

Prove that $f$ is integrable on $[0,1]$ and $\int_0^1fd\mu=\lim_{n\to\infty}\int_0^1f_nd\mu$.

How to prove this? I confronted with some difficulties.

My idea: Let \begin{eqnarray*} f_n^N=\max \{\min\{f_n,N\},-N\},\\ f^N=\max \{\min\{f,N\},-N\}. \end{eqnarray*} Then for any $x\in[0,1]$, $|f_n(x)|\leq N$, $|f(x)|\leq N$. By Lebesgue Dominated Convergence Theorem, for each $N\in\mathbb{N}$, \begin{eqnarray*} \lim_{n\to\infty}\int f_n^Nd\mu=\int\lim_{n\to\infty}f_n^Nd\mu =\int f^Nd\mu. \end{eqnarray*} Since $|f_n^N|\leq |f|$, by Lebesgue Dominated Convergence Theorem, for each $n\in\mathbb{N}$, \begin{eqnarray*} \lim_{N\to\infty}\int f_n^Nd\mu=\int\lim_{N\to\infty}f_n^Nd\mu=\int f_nd\mu. \end{eqnarray*} By Levi's monotone convergence theorem, \begin{eqnarray*} \int fd\mu&=&\int f^+d\mu-\int f^-d\mu\\ &=&\int\lim_{N\to\infty}f^{N+}d\mu-\int\lim_{N\to\infty}f^{N-}d\mu\\ &=&\lim_{N\to\infty}\int f^{N+}d\mu-\lim_{N\to\infty}\int f^{N-}d\mu\\ &=&\lim_{N\to\infty}\int f^Nd\mu. \end{eqnarray*} Letting $N\to\infty$ in the first equality,

\begin{eqnarray*} \int fd\mu&=&\lim_{N\to\infty}\int f^N d\mu\\ &=&\lim_{N\to\infty}\lim_{n\to\infty}\int f^N_n d\mu\\ &=&\lim_{n\to\infty}\lim_{N\to\infty}\int f^N_n d\mu \\ & &\text{ (need to prove the order of two limits can be exchanged)}\\ &=& \lim_{n\to\infty}\int f_nd\mu. \end{eqnarray*}

To prove that the order of the two limits can be reversed, we only need to prove:

(1). $\lim_{n\to\infty,N\to\infty}f^N_nd\mu$ exists;

(2). for any $n\in\mathbb{N}$, $\lim \int f^N_nd\mu$ exists;

(3). for any $N\in\mathbb{N}$, $\lim_{n\to\infty}\int f^N_n d\mu$ exists.

(2) is obtained by Lebesgue dominated convergence theorem.

(3) is obtained by Egoroff theorem: for any $\delta>0$, there exists $E\subseteq [0,1]$, $\mu(E)<\delta$ such that $f_n^N\to^n f^N$ uniformly on $[0,1]\setminus E$. Hence for any $\epsilon >0$, there exists $n_0$ such that for any $n\geq n_0$, $|f^N_n-f^N|<\epsilon$. Thus \begin{eqnarray*} |\int_{[0,1]\setminus E} f^N_n d\mu-\int_{[0,1]\setminus E} f^N d\mu|<\epsilon. \end{eqnarray*} Let $N$ be fixed. Choose $\epsilon=2N\delta$. Then \begin{eqnarray*} |\int_{E} f^N_n d\mu-\int_{E} f^N d\mu|\leq \int_E |f^N_n-f^N|d\mu\leq 2N\mu(E)<\epsilon. \end{eqnarray*} Hence for all $n\geq n_0$, \begin{eqnarray*} |\int f^N_n d\mu-\int f^N d\mu| &\leq&|\int_{E} f^N_n d\mu-\int_{E} f^N d\mu|+|\int_{[0,1]\setminus E} f^N_n d\mu-\int_{[0,1]\setminus E} f^N d\mu|\\ &<& 2\epsilon. \end{eqnarray*} Since $\delta>0$ is arbitrary, $\epsilon=2N\delta$ is also arbitrary. Hence we obtain (3).

But I do not know how to prove (1).

Until now, I have not used the given conditions (a), (b). How to prove the double limit exists by applying (a) and (b)? Without (a) or (b), the double limit in (1) may not exist?

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  • $\begingroup$ This may help. $\endgroup$ – David Mitra May 27 '14 at 13:28
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It might be eaiser to use the following argumentation:

First of all, it follows from Fatou's lemma that

$$\int |f| \, d\mu = \int \liminf_{n \to \infty} |f_n| \, d\mu \leq \liminf_{n \to \infty} \int |f_n| \, d\mu < \infty$$

by a). Hence, $f \in L^1$. From this and b) it follows that we can find for any $\varepsilon>0$ some $\delta>0$ such that $\mu(E)<\delta$ implies

$$\int_E |f_n| \, d\mu < \varepsilon, \quad n \in \mathbb{N}, \qquad \int_E |f| \, d\mu< \varepsilon. \tag{1}$$

Now fix $\varepsilon>0$. As $f_n \to f$ almost everywhere, we know that $\mu(|f_n-f|> \varepsilon) \to 0$ as $n \to \infty$. In particular, if we choose $n$ sufficiently large, then $\mu(E)<\delta$ for $E := \{|f_n-f|>\varepsilon\}$. Hence,

$$\begin{align*} \int |f_n-f| \, d\mu &= \int_E |f_n-f| \, d\mu + \int_{E^c} \underbrace{|f_n-f|}_{\leq \varepsilon} \, d\mu \\ &\leq \int_E |f_n| \, d\mu+ \int_E |f| \, d\mu+ \varepsilon \mu([0,1]) \\ &\stackrel{(1)}{\leq} \varepsilon (2+\mu([0,1]).\end{align*}$$

This finishes the proof.

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