Friday, 31 May 2013

lebesgue integral - Exercise in finite measure space.



Let (X,μ) be a finite measure space and f:X[0,] be a measurable function. Define a function φ:R+R+ by φ(r)=(1μ(X)Xfrdμ)1r. Prove that φ(r)φ(s) if $r

Answer



We have
φ(r)=(1μ(X)Xfr dμ)1r=(Xfr 1μ(X) dμ)1r.
Now, by Hölder's inequality, in general, for measurable functions F,G and conjugate exponents p,q, that is, 1p+1q=1 where $1ϕ(r)=(XFG dμ)1r
as desired.


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