a special measurable function

**mgarson** · September 4th 2011, 11:51 AM

Hi,

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

How do I prove the following:

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

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

(i) For this one I'm thinking that the condition on $h$ implies that $h$ is constant a.e., say $h=C$ a.e. so
$\lim_{k\rightarrow\infty}\int_a^bh_k(x)dx=C\int_a^ b1dx=m((a,b))C=(b-a)\int_0^1h(x)dx$ .
But if that is true... the the $\lim_k$ 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...

**girdav** · September 4th 2011, 12:14 PM

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

**Opalg** · September 4th 2011, 12:23 PM

Originally Posted by mgarson

Hi,

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

How do I prove the following:

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

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

(i) For this one I'm thinking that the condition on $h$ implies that $h$ is constant a.e., say $h=C$ 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 $h(x) = \sin(2\pi x).$

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 $\int_a^bh_k(x)\,dx,$ divide the interval [a,b] into subintervals of length 1/k and make the substitution $y=kx$ to show that $\int h(kx)\,dx$ over each of these subintervals is equal to $\frac1k\int_0^1h(x)dx.$ 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 $k\to\infty,$ 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 $h_k$ 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 $u\in(a,b],$ $\int_a^uf(x)\,dx = (u-a)\int_0^1h(x)dx.$ 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.)

**mgarson** · September 6th 2011, 08:52 AM

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

All I'm a bit concerned about is the follwoing equality:

Originally Posted by Opalg

$\int_a^uf(x)\,dx = (u-a)\int_0^1h(x)dx.$

what we´re using here is (i), i.e $\lim_{n\rightarrow\infty}\int_a^uh_{k_n}(x)dx=(u-a)\int_0^1h(x)dx$ . But if the same proof as in (i) is to hold we need to have $k_n\rightarrow \infty$ as $n\rightarrow\infty$ . Does this really need to be true? What if $k_n\rightarrow A<\infty$ as $n\rightarrow\infty$ ?

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