What happens if ?
Hi,
I have a bdd Lebesgue measurable function on such that h(x)=h(x+1) a.e. on . Define .
How do I prove the following:
(i)
(ii) If isn't constant a.e. then there will be no subseq that will converge -a.e. on .
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My attempts:
(i) For this one I'm thinking that the condition on implies that is constant a.e., say a.e. so
.
But if that is true... the the 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...
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
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 divide the interval [a,b] into subintervals of length 1/k and make the substitution to show that over each of these subintervals is equal to 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 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 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 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.)
Ok, so knowing that , where is some constant, should imply that is constatnt a.e.
So, which then gives me a.e. and therefore also a.e.
All I'm a bit concerned about is the follwoing equality:
what we´re using here is (i), i.e . But if the same proof as in (i) is to hold we need to have as . Does this really need to be true? What if as ?