Monday, 20 May 2019

real analysis - What is so special about the Lebesgue-Stieltjes measure




A measure λ:B(Rn)¯R0 that is associated with a monotone increasing and right-side continuous function F is called a Lebesgue-Stieltjes measure. But I am wondering, why it is not true that every measure λ:B(Rn)¯R0 is a Lebesgue-Stieltjes measure?


Answer



μ being a lebesgue-stiltjes measure with corresponding function F implies that μ((a,b])=F(b)F(a).



Now take the (rather silly) measure μ(X)={if 0X1if 0X1X0otherwise.



We'd need to have F(x)= for x0 and F(x)=0 for x<0 to have μ((a,b])=F(b)F(a) for a<0, b0. But then μ((0,2])=F(2)F(0)=


which is





  1. meaningless, and

  2. surely not the same as 1, which is the actual measure of (0,2].



Note that a measure doesn't necessarily need to have infinite point weights (i.e., x for which μ({x})= to cause trouble. Here's another measure on B(R) which isn't a lebesgue-stiltjes measure μ(X)=nN,1nX1n.


For every ϵ>0, μ((0,ϵ])=, which again would require F(x)= for x>0, and again that conflicts the requirement that μ((a,b])=F(b)F(a)< for 0<ab. Note that this measure μ is even σ-finite! You can write R as the countable union R=(,0]=A(1,)=BnN(1n+1,1n]=Cn


and all the sets have finite measure (μ(A)=μ(B)=0, μ(Cn)=1n).



You do have that all finite (i.e., not just σ-finite, but fully finite) measures on B(R) are lebesgue-stiltjes measures, however. This is important, for example, for probability theory, because it allows you to assume that every random variable on R has a cumulative distribution function (CDF), which is simply the function F.


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