Measurable Function and Sequence of Real Numbers

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

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.

Re: Measurable Function and Sequence of Real Numbers

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

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

If you now let n vary, you should be able to show that given N, there is a subset $\displaystyle E_N$ of E with measure less than $\displaystyle 2^{-N+1}$ with the property that if $\displaystyle x\in E\setminus E_N$ then $\displaystyle |\alpha_nf_n(x)|<2^{-n}$ for all $\displaystyle n\geqslant N.$ It follows that $\displaystyle \alpha_nf_n(x)\to0$ except on the null set $\displaystyle \textstyle\bigcap_N E_N.$

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.

Re: Measurable Function and Sequence of Real Numbers

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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 $\displaystyle f_n$ 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.