# Does Radon-Nikodym imply Riesz Representation Theorem?

In Axler's Linear Algebra Done Right we have the theorem

6.42: (Riesz Representation Theorem) Suppose $$V$$ is a finite dimensional inner product space and $$\phi$$ is a linear functional on $$V$$. Then there is a unique vector $$u \in V$$ such that $$\phi(v) = \langle v, u\rangle$$ for every $$v \in V$$

I'm currently learning measure theory and have came across Radon-Nikodym

(Radon-Nikodym) Consider a measurable space $$(X,\mathcal{M})$$ on which two $$\sigma$$-finite signed measures $$\mu,\nu$$ are defined such that $$\nu << \mu$$ ($$\nu$$ is absolutely continuous with respect to $$\mu$$) then there is a $$\mu$$-integrable function $$f: X \to \mathbb{R}$$ such that $$\nu(E) = \int_E f d\mu$$ for every $$E \in \mathcal{M}$$ and any other function $$g$$ satisfying this is equal to $$f$$ almost everywhere with respect to $$\mu$$.

These two theorems seems very similar. Is it possible to go from Radon-Nikodym and get Riesz Representation? The integral in Radon-Nikodym "acts" like the inner product in Riesz Representation, the function $$f$$ "acts" like the vector $$u$$ in Riesz, and the signed measure $$\nu$$ acts like the linear functional $$\phi$$.

I am inclined to think that somehow we can recover Riesz from Radon-Nikodym. To start, we would need to somehow get a $$\sigma$$-algebra, $$\mathcal{M}$$ on $$V$$ so that $$(V, \mathcal{M})$$ is a measurable space. This has to be a very particular $$\sigma$$-algebra so that somehow the integral can be reduced to the inner product on $$V$$. We would also need to show that the linear functional is absolutely continuous with respect to the inner product.

So is it possible to recover Riesz from Radon-Nikodym? If so, how? If not, what's the issue?

• if $\phi$ is a linear functional in $L^p$ I know that this that your claim is true... define $\lambda(E)=\phi(\chi_E)$, if $\phi$ is positive we can (generalize after) show that $\lambda$ is a measure and if the set $E$ has measure $\mu(E)=0$ than $\lambda(E)=0$ because $\phi$ is linear... then use Radon nikodym and you show that every linear functional of $L^p$ is $\phi(f)=\int fg d\mu$ for some $g\in L^q$ – Robson Nov 18 '18 at 22:20

## 1 Answer

I am not sure about your specific claim, but certainly there is a connection in the "opposite" direction. That is, a general version of the Riesz Representation Theorem is used to prove the Radon-Nikodym Theorem. See "von Neumann's proof" of the Radon-Nikodym Theorem (for instance, Section 8.1, in particular Theorem 8.1.3, of Daniel Stroock's "Essentials of Integration Theory for Analysis," DOI:10.1007/978-1-4614-1135-2, or a neat summary at https://blameitontheanalyst.wordpress.com/2010/06/23/von-neumanns-proof-of-radon-nikodym/).

• I think this is a good answer though. – user370967 Jan 30 at 15:52
• Although not the exact thing I was looking for, this is a very good answer. Thank you. – carsandpulsars Feb 4 at 4:10