so I have been working on this problem and I want to make sure I am understanding the conclusion fully. So I have the following scenario:

Not part of the actual question, but relevant. Consider the measure space $([0,1],\mathcal{M},m),$ where $\mathcal{M}$ is the set of Lebesgue measurable sets contained in $[0,1],$ and $m$ is the Lebesgue measure.

Let $f$ be some function s.t. $$\int_a^b f dm =b-a,$$ and define $$\int_A 1d\mu=\mu(A):=\int_A f dm.$$ Thus we have for any $[a,b]\subseteq[0,1]$ $$\mu([a,b])=b-a.$$

I want to show $f=1$ [m] a.e., and I think good way to go about this would be to show $m=\mu.$

Now if we consider the set $L,$ which is where the Lebesgue measure and $\mu$ agree we have that $I,$ the set of intervals is contained in the former, i.e. $I\subseteq L.$ Now I is clearly a $\pi$ system and $L$ is a $\lambda$ system, which means by Dynkin's Theorem $\sigma(I)=\mathcal{B}\subseteq L,$ where $\mathcal{B}$ is the Borel sets on $[0,1].$

Now this would give me that $\mu=m,$ but only on the Borel sets, and hence I can't say that $m=\mu$ on $\mathcal{M}.$ Is this correct or am I missing something?

I also see that $f$ is the Radon-Nikodym derivative, which means it is unique and we know $f=1$ works (on intervals). I don't think this gives me the result fully though.

Thanks for any help.

  • 2
    $\begingroup$ Lebesgue measurable set can be written as the union of a Borel set and null set. By definition, $\mu(N) = \int_N f \, dm = 0$ when $N$ is a $m$-null set. Then is it enough to show what you want ? $\endgroup$ – Chival Jul 9 '15 at 7:49

You can directly appeal to the Lebesgue differentiation theorem. For m-a.e. $x\in [0,1]$, \begin{align*} \lim_{h\rightarrow 0^{+}}\dfrac{1}{2h}\int_{x-h}^{x+h}f\mathrm{d}m=\lim_{h\rightarrow 0^{+}}1=1 \end{align*}

Chival gave you another way of answering your question. The Lebesgue $\sigma$-algebra is the completion of the Borel $\sigma$-algebra, so every Lebesgue measurable set $E$ is a union of a Borel set $A$ and a (Lebesgue) null set $N$. So $E\setminus A=N\setminus A$, which is a null set. Whence,


  • $\begingroup$ Thanks for the additional solution. I had thought of using differentiation in some way, but I wasn't sure if it was fair game. I am working through a prelim pack and not all the teachers cover differentiation. So even though mine did, I try to solve some of the problems other ways if I can. Extra practice. $\endgroup$ – Scott Jul 16 '15 at 6:26

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