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Given that $(X,\mathcal{F},\mu)$ is a finite measure space and $\{f_k\}$ is a sequence of finite-valued measurable functions, such that for any $\varepsilon > 0$

$$ \lim_{n \rightarrow \infty} \mu(\{x \in X : \sup_{k\geq n} |f_k(x)| \geq \varepsilon\}) =0 .$$

I need to show that $\lim_{k \rightarrow \infty} f_k(x) = 0$ $\mu$-a.e on $X$.

The equality is easy to show. But I am stuck at showing explicitly that the limit $\lim_{k \rightarrow \infty} f_k(x)$ exists $\mu$-a.e on $X$.

Any help/comment is greatly appreciated !

Note that by "finite-valued function", it is only meant that the function does not take the values $+\infty$ or $-\infty$ on $X$.

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  • $\begingroup$ Maybe I've misunderstood your question since I don't know how you could have shown the equality without proving the limit exists a fortiori. But in any case, suppose the limit doesn't exist on a set of positive measure, $A$. Then there exists $\epsilon >0 $ such that $\sup_{k \geq n} |f_k(x)|> \epsilon$ for all $x \in A$ and all $n$. This is a contradiction since $A$ has positive measure. $\endgroup$ – Tim kinsella Feb 7 '14 at 3:24
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Fix $\varepsilon>0$. Let $g_n=\sup\{|f_k|:\ k\geq n\}$. Our assumption is that $\lim_n\mu\{g_n\geq\varepsilon\}=0$. We can write this as $\lim_n\mu\{g_n<\varepsilon\}=\mu(X)$.

The sequence $\{g_n\}$ is decreasing and positive; in particular it is convergent. So the sets $\{g_n<\varepsilon\}$ form an increasing sequence. Then, using continuity of the measure, $$ \mu(\bigcup_n\{g_n<\varepsilon\})=\lim_n\mu(\{g_n<\varepsilon\})=\mu(X). $$ That is, up to a null-set, $X=\bigcup_n\{g_n<\varepsilon\}$. In other words, $\lim_ng_n<\varepsilon$ a.e. As $\varepsilon$ was arbitrary, $\lim_ng_n=0$. This means that $$ \limsup_k|f_k|=0\ \text{ a.e. }, $$ which implies $\lim_k|f_k|=0$ a.e.

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Let $$A_{n,\epsilon} = \{|f_n| \geq \epsilon \}, $$ $$A_\epsilon = \limsup_n A_{n,\epsilon} = \bigcap_{n=1}^\infty\bigcup_{k=n}^{\infty}A_{k,\epsilon}.$$ As $\bigcup_{k=n}^{\infty}A_{k,\epsilon}$ is decreasing with intersection over $n$ equal to $ A_\epsilon $, we have: $$\lim_n \mu\left(\bigcup_{k=n}^{\infty}A_{k,\epsilon} \right) = \mu\left(A_\epsilon \right). $$ Now: $$ \left\{\lim_n f_n \not= 0\right\} = \bigcup_{\epsilon>0} A_\epsilon =\bigcup_{m=1}^{\infty} A_{1/m}.$$ Last equality is needed to show the measurability of this set.

So:

$\lim_n f_n = 0$ a.e. if and only if $$\mu\left(A_\epsilon \right)=0$$ for all $\epsilon > 0$ if and only if $$\lim_n \mu\left(\bigcup_{k=n}^{\infty}A_{k,\epsilon} \right) =0$$ for all $\epsilon > 0$.

Finally, note that $$ \bigcup_{k=n}^{\infty}A_{k,\epsilon} \subseteq \left\{\sup_{k\geq n} |f_k|\geq \epsilon\right\}.$$

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