# Thread: a special measurable function

1. ## a special measurable function

Hi,

I have a bdd Lebesgue measurable function $\displaystyle h$ on $\displaystyle \mathbb{R}$ such that h(x)=h(x+1) a.e. on $\displaystyle \mathbb{R}$. Define $\displaystyle h_k(x)=h(kx)$.

How do I prove the following:

(i) $\displaystyle \lim_{k\rightarrow\infty}\int_a^bh_k(x)dx=(b-a)\int_0^1h(x)dx$

(ii) If $\displaystyle h$ isn't constant a.e. then there will be no subseq that will converge $\displaystyle m$-a.e. on $\displaystyle $a,b$$.
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My attempts:

(i) For this one I'm thinking that the condition on $\displaystyle h$ implies that $\displaystyle h$ is constant a.e., say $\displaystyle h=C$ a.e. so
$\displaystyle \lim_{k\rightarrow\infty}\int_a^bh_k(x)dx=C\int_a^ b1dx=m((a,b))C=(b-a)\int_0^1h(x)dx$.
But if that is true... the the $\displaystyle \lim_k$ is of no use.

(ii) Here I'm thinking a contradiction, i.e. assuming that there is such a sub.seq. but I get it to work...

2. ## Re: a special measurable function

What happens if $\displaystyle h(x)=\sin (2\pi x)$?

3. ## Re: a special measurable function

Originally Posted by mgarson
Hi,

I have a bdd Lebesgue measurable function $\displaystyle h$ on $\displaystyle \mathbb{R}$ such that h(x)=h(x+1) a.e. on $\displaystyle \mathbb{R}$. Define $\displaystyle h_k(x)=h(kx)$.

How do I prove the following:

(i) $\displaystyle \lim_{k\rightarrow\infty}\int_a^bh_k(x)dx=(b-a)\int_0^1h(x)dx$

(ii) If $\displaystyle h$ isn't constant a.e. then there will be no subseq that will converge $\displaystyle m$-a.e. on $\displaystyle $a,b$$.
-----------------------------------------------------------
My attempts:

(i) For this one I'm thinking that the condition on $\displaystyle h$ implies that $\displaystyle h$ is constant a.e., say $\displaystyle h=C$ a.e.
The condition on h says that it is periodic, with period 1. That certainly does not imply that h is constant a.e. For example, you could have $\displaystyle h(x) = \sin(2\pi x).$

To prove (i), you could start by using the periodicity of h to show that the integral of h over any interval of length 1 is equal to the integral from 0 to 1. Then to find $\displaystyle \int_a^bh_k(x)\,dx,$ divide the interval [a,b] into subintervals of length 1/k and make the substitution $\displaystyle y=kx$ to show that $\displaystyle \int h(kx)\,dx$ over each of these subintervals is equal to $\displaystyle \frac1k\int_0^1h(x)dx.$ The only snag is that b–a may not be an exact multiple of 1/k, so there may be a little subinterval left over at the end. As $\displaystyle k\to\infty,$ the contribution of the integral over that subinterval will become negligible.

For (ii), I would start as you suggest, suppose that there is a subsequence of the functions $\displaystyle h_k$ that converges a.e. on [a,b] to a limit function f. Use one of the Lebesgue convergence theorems, together with the result of (i), to deduce that for each $\displaystyle u\in(a,b],$ $\displaystyle \int_a^uf(x)\,dx = (u-a)\int_0^1h(x)dx.$ Then differentiate both sides to deduce that f is constant a.e. You still have to deduce that this implies that h is constant a.e., and I must admit I haven't thought why that follows. (But I think it should.)

4. ## Re: a special measurable function

Ok, so knowing that $\displaystyle f=C$, where $\displaystyle C$ is some constant, should imply that $\displaystyle f$ is constatnt a.e.
So, $\displaystyle \int_0^1h(x)dx=C$ which then gives me $\displaystyle h_k-C=0$ a.e. and therefore also $\displaystyle h-C=0$ a.e.

All I'm a bit concerned about is the follwoing equality:
Originally Posted by Opalg
$\displaystyle \int_a^uf(x)\,dx = (u-a)\int_0^1h(x)dx.$
what we´re using here is (i), i.e $\displaystyle \lim_{n\rightarrow\infty}\int_a^uh_{k_n}(x)dx=(u-a)\int_0^1h(x)dx$. But if the same proof as in (i) is to hold we need to have $\displaystyle k_n\rightarrow \infty$ as $\displaystyle n\rightarrow\infty$. Does this really need to be true? What if $\displaystyle k_n\rightarrow A<\infty$ as $\displaystyle n\rightarrow\infty$?