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 $$g$$?

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.