A Question on Lebesgue Decomposition

[haljordan45]

Homework Statement

Let μ be the counting measure and m be the Lebesgue measure. Then show that on the interval [0,1] m has no Lebesgue decomposition with respect to μ.

Homework Equations

If such a decomposition exists, then the following holds true where X is the whole space, E is a subset of X, and X_s is the singular subset of the space:

1. m=m_a+m_s where m_a is absolutely continuous and m_s is singular
2. m_a(E)=∫_Efdμ
3. m_s(X-X_s)=μ(X_s)=0

The Attempt at a Solution

I know how to show that μ has no Lebesgue decomposition with respect to m, but can't seem to figure out this direction. I'm assuming that I need to pick a set E that contradicts 3 above, but I'm at a loss.

 

[mighty2000]

To show that m has no Lebesgue decomposition with respect to μ, we need to find a set E such that 3 above is not satisfied. One way to do this is to choose a set E that is not measurable with respect to μ. Let E be a non-measurable subset of [0,1]. Then, by definition of the Lebesgue measure, m(E) = 0 since E is not measurable. However, if we assume that m has a Lebesgue decomposition with respect to μ, then we must have m(E) = ma(E) + ms(E). Since m(E) = 0, this implies that ma(E) = -ms(E). But since ma is absolutely continuous with respect to μ, we must have ma(E) = 0. This would imply that ms(E) = 0, which contradicts the fact that E is a non-measurable set. Therefore, we have shown that on the interval [0,1], m has no Lebesgue decomposition with respect to μ.