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I'm new and I have a question I need help in solving. This isn't homework. The question is as follows:

Let $(X,M,\mu)$ be a finite measure space and suppose $f_n$ is a sequence of integrable functions on $X$. Suppose also, that to each $\epsilon \gt 0$ there is an integer $N\gt 0$ such that $\int_E |f_n|~d\mu \lt \epsilon$ for all measurable sets $E$ satisfying $\mu(E)\lt 1/N$ and all integers $n$ satisfying $n \gt N$. If $f_n\rightarrow f$ a.e. $[\mu]$ on $X$, I want to show that $f_n\rightarrow f$ in $L^1$-norm.

Thanks very much for your help.

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A slight modification of the proof on pgs 89-90 (Theorem 5.47) of this should work.… – David Mitra Nov 28 '11 at 19:12
up vote 3 down vote accepted

Your question is a particular version of Vitali's Convergence Theorem:

Theorem: Let $1\le p<\infty$, $(X,\Sigma,\mu)$ be a measure space, and let $\{f_n\}$ be a sequence in $L_p(\mu)$. Then $f_n$ converges to a measurable $f$ in $L_p$ if and only if the following conditions hold:

  1. $f_n$ converges in measure to $f$.

  2. $\{|f_n|^p\}$ is uniformly integrable. That is, for every $\epsilon>0$, there is a $\delta>0$ such that for any set $E$ of measure less than $\delta$: $$ \int_E |f_n|^p<\epsilon $$ for each $n$.

  3. For every $\epsilon>0$, there exists a set $E$ of finite measure such that $\int_{E^C}|f_n|^p<\epsilon$ for all $n$.

With your conditions:

Condition 3. is satisfied trivially, since you have a finite measure.

Condition 1. is satisfied since, in finite measure spaces, a.e convergence implies convergence in measure.

Condition 2. is satisfied from your conditions and the fact that finite sets in $L_1$ are uniformly integrable.

So, if you like, you can just appeal to Vitali.

Some hints if you wish to prove your result from scratch:

First show that $f\in L_1$. Then show $f_n$ converges to $f$ in $L_1$.

For both parts, the strategy is to appeal to Egorov to find a set $E$ of small measure such that $f_n$ converges to $f$ uniformly off $E$. Do this in such a way that on the set $E$, you can take advantage of uniform integrability.

The details of this proof can be found here on pages 89-90.

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Let $\epsilon>0$. Choose $N$ such that for $n>N$, $\mu(E) < 1/N$ \implies $\int_E |f_n| < \epsilon$ and $\int_E |f| < \epsilon$. This can be done since $|f|\,d\mu << \mu$.

Since the measure space is finite, $f_n \rightarrow f$ in measure. Now choose $N_2 > N$ such that $n>N_2$ implies $\mu(|f_n-f|>\epsilon/(\mu(X)+1)) < 1/N$.

Then, for $n>N_2$, and setting $E_n \equiv (|f_n-f|>\epsilon/(\mu(X)+1))$, we have that $\mu(E_n) < 1/N$:

$\int |f_n - f| = \int_{E_n} |f_n-f| + \int_{E_n^c} |f_n - f| \leq \int_{E_n} |f_n| + \int_{E_n} |f| + \int_{X} \frac{\epsilon}{\mu(X)+1} < 3\epsilon$.

Therefore, for any $\delta>0$, setting $\epsilon = \delta/3>0$ there exists an $N$ such that $n>N$ imply that $\int |f_n - f| \,d\mu < 3\epsilon = \delta$. Thus $f_n \rightarrow f$ in $L^1$

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