Sequence of Lebesgue measurable functions

**Diamondlance** · November 14th, 2008, 10:42 PM

I want to find a sequence $f_n(x) \in L^+$ such that $f_n$ is defined on (0,1) for all natural numbers $n$ , we have $lim_{n\rightarrow\infty}\int f_n(x)dm=0$ (m=Lebesgue measure), and $lim_{n\rightarrow\infty}f_n(x)$ does not exist for any x in (0,1).

My initial idea, which I know does not work, was to define something like this. For n odd, let $f_n(x)$ be $\displaystyle\frac{2}{n+1}$ for irrational x, and 0 otherwise. For n even, let $f_n(x)$ be 0 for irrational x and 1 for rational x. That satisfies the first condition about the limit of the integral, but does not satisfy the second condition since the limit of $f_n(x)$ is 0 on the irrationals.

Even though I know this doesn't work, I was wondering if someone here would know of a way I could think about tweaking this to work, or am I barking up the wrong tree, so to speak?

**Opalg** · November 14th, 2008, 11:51 PM

Originally Posted by Diamondlance

I want to find a sequence $f_n(x) \in L^+$ such that $f_n$ is defined on (0,1) for all natural numbers $n$ , we have $\lim_{n\rightarrow\infty}\int f_n(x)dm=0$ (m=Lebesgue measure), and $lim_{n\rightarrow\infty}f_n(x)$ does not exist for any x in (0,1).

The idea is to construct a sequence of functions that are 0 on most of the interval but jump up to 1 on a small subinterval that sweeps across the whole interval.

Specifically, given any natural number n, let $2^q$ be the largest power of 2 that is $\leqslant n$ , so that $n = 2^q + r$ , where $0\leqslant r<2^q$ . Then define $f_n(x) = \begin{cases}1&\text{if }r/2^q\leqslant x<(r+1)/2^q,\\ 0&\text{otherwise.}\end{cases}$

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