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I came across the following question asked in a prelims of UW.

Let $(X,F,\mu)$ be a finite measure space, and $\lambda$ be a finitely-additive, non-negative, real valued set function on $F$ such that $$ \forall \epsilon >0, \exists \delta >0 : \mu(E)<\delta \implies \lambda(E)<\epsilon.$$ Show that there exists $g\in L^1(\mu)$ such that $$\lambda(E)=\int_E g d\mu.$$

The hypothesis clearly implies that $\lambda$ is absolutely continuous wrt $\mu.$ One therefore expects to invoke the Radon-Nikodym theorem. But, I know the Radon-Nikudym theorem to be tre only when $\lambda$ is also given to be a measure (I.e. countably additive). So, I tried mimicking the proof of Radon-Nikodym theorem, but I have a doubt in one step.

We define $d\mu_1= d\lambda + d\mu$. And, look at the linear functional $$f \to \int_X f d\lambda.$$ I can show that this is a bounded linear functional on $L^2(\mu_1).$ And, therefore we invoke Riesz representation theorem (?) to obtain a function $h\in L^2(\mu_1)$ such that $$\int_X f d\lambda=\int_X fh d\mu_1.$$ From this we obtain that $$\int_X f(1-h) d\lambda = \int_X fhd\mu.$$

After this step the usual procedure shows that $0\leq h\leq 1$. And, the set on which $g=1$ is of measure 0 wrt $\mu$ and therefore wrt $\lambda.$ Replacing $f$ by $(1+h+...+h^n)f$ (We can do this because $h$ is bounded) and taking limit as $n\to \infty$ we obtain that $$\int_X f d\lambda = \int_X f \frac{h}{1-h}d\mu$$ which fives us the desired conclusion.

I understand that $\mu$ being finite measure is not crucial but is helpful and simplifies things a bit. My confusion is that $d\mu_1$ is not an honest “measure” because it is only finitely additive. In this situation can we invoke RRT? Secondly, should the assumption of finite additivity on $\lambda$ make our life easier? I mean is it possible to write a different and hopefully simpler proof for the existence of $g$?

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If $(E_n)$ decreases to $0$ then $\mu(E_n) \to 0$. From the given hypothesis it follows that $\lambda (E_n) \to 0$. Any finitely additive measure with this property is automatically countably additive, so the usual Radon Nikodym Theorem is applicable.

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  • $\begingroup$ Ohhh! Yes, I missed that. This was really pretty. Thanks a lot. $\endgroup$ – WhoKnowsWho Jul 2 at 10:35

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