A measure λ:B(Rn)→¯R≥0 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)→¯R≥0 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 0∈X1if 0∉X, 1∈X0otherwise.
We'd need to have F(x)=∞ for x≥0 and F(x)=0 for x<0 to have μ((a,b])=F(b)−F(a) for a<0, b≥0. But then μ((0,2])=F(2)−F(0)=∞−∞
which is
- meaningless, and
- 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)=∑n∈N,1n∈X1n.
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<a≤b. Note that this measure μ is even σ-finite! You can write R as the countable union R=(−∞,0]⏟=A∪(1,∞)⏟=B∪⋃n∈N(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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