# Measurable Function and Sequence of Real Numbers

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• Sep 26th 2011, 09:37 AM
Markeur
Measurable Function and Sequence of Real Numbers
Let $E$ be a Lebesgue measurable subset of $\mathbb{R}$ with $m(E) < \infty$ and let $(f_n)$ be a sequence of real-valued Lebesgue measurable functions on $E$. Prove or disprove that there exists a sequence $(\alpha_n)$ of positive real numbers such that $lim_{n \to \infty} \alpha_n f_n = 0$ almost everywhere on $E$.

I suppose the statement is true? So far can't think of any counterexample. If it's true, how do I go about starting the proof?

Thanks in advance.
• Sep 27th 2011, 01:49 AM
Opalg
Re: Measurable Function and Sequence of Real Numbers
Quote:

Originally Posted by Markeur
Let $E$ be a Lebesgue measurable subset of $\mathbb{R}$ with $m(E) < \infty$ and let $(f_n)$ be a sequence of real-valued Lebesgue measurable functions on $E$. Prove or disprove that there exists a sequence $(\alpha_n)$ of positive real numbers such that $lim_{n \to \infty} \alpha_n f_n = 0$ almost everywhere on $E$.

I suppose the statement is true? So far can't think of any counterexample. If it's true, how do I go about starting the proof?

Thanks in advance.

For fixed n, the union of the sets $S_k = \{x\in E:|f_n(x)|>k\}\ \ (k\in\mathbb{N})$ is the whole of E. So there exists K such that $m(S_K)>m(E)-2^{-n}.$ Let $\alpha_n = 1/(2^nK).$ Then $|\alpha_nf_n(x)|<2^{-n}$ except on a subset of E of measure less than $2^{-n}.$

If you now let n vary, you should be able to show that given N, there is a subset $E_N$ of E with measure less than $2^{-N+1}$ with the property that if $x\in E\setminus E_N$ then $|\alpha_nf_n(x)|<2^{-n}$ for all $n\geqslant N.$ It follows that $\alpha_nf_n(x)\to0$ except on the null set $\textstyle\bigcap_N E_N.$
• Sep 27th 2011, 09:10 AM
Markeur
Re: Measurable Function and Sequence of Real Numbers
Hey thanks very much! Anyway I would like to know why we need to establish S_k = {x \in E : |f_n(x)| > k}. Is it because (f_n) is a sequence of real-valued Lebesgue measurable functions without any other condition given?

Thanks in advance.
• Sep 27th 2011, 10:21 AM
Opalg
Re: Measurable Function and Sequence of Real Numbers
Quote:

Originally Posted by Markeur
Hey thanks very much! Anyway I would like to know why we need to establish S_k = {x \in E : |f_n(x)| > k}. Is it because (f_n) is a sequence of real-valued Lebesgue measurable functions without any other condition given?

If you knew that the functions were bounded then the result would be easy to prove. But a measurable function, even on a set of finite measure, need not be bounded. So the idea is to use the measurability condition to find a "large" subset of E on which $f_n$ is bounded. But the details of the proof are quite delicate, and I have had to edit my previous comment a couple of times to get an argument that (I think!) works.