Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $X$.

share|cite|improve this question
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. – Tim kinsella Feb 7 '14 at 3:24
up vote 0 down vote accepted

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.

share|cite|improve this answer

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.


$\lim_n f_n = 0$ 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 $$ \bigcup_{k=n}^{\infty}A_{k,\epsilon} \subseteq \left\{\sup_{k\geq n} |f_k|\geq \epsilon\right\}.$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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