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Let $(X,\mu)$ be a measure space and $\mu(X)< + \infty$, $\phi$ be a bounded linear functor on $L^1(\mu)$. Prove that there exists a positive measure $\lambda$ on $X$ such that $\phi(f) = \int_X f d\lambda$ for any $f \in L^1(\mu)$.

This appears like Riesz representation theorem to me. But I don't know how to do it.

Thank you very much.

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This is the abstract version of the Riesz representation theorem. Check out this wikibook entry for a proof on $C_c(X)$ (which is dense in $L^1(X,d\mu)$) or for the full picture see Rudin "Real and Complex Analysis" or Lax "Functional Analysis" –  icurays1 Nov 22 '12 at 2:42
    
@icurays1 What happens if there is not a topological structure on $X$? –  Davide Giraudo Nov 23 '12 at 8:44

1 Answer 1

If $\lambda$ works, then $\lambda(A)=\phi(\chi_A)$ for all $A$ measurable (as the measure is finite, $\chi_A\in L^1(\mu)$).

So we define $\lambda(A):=\phi(\chi_A)\geqslant 0$. It defines a set function, with $\lambda(\emptyset)=0$. $\sigma$-additivity is a consequence of monotone convergence theorem.

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Out of curiosity: How exactly are you suggesting to apply the monotone convergence theorem? Do you mean that the characteristic functions of $F_n$ converge in $L^1$ to the characteristic function of $\bigcup F_n$ or is there a slicker way? Further, a minor point: there seems to be an assumption on positivity of the functional missing in the question. –  commenter Nov 22 '12 at 22:59
    
I don't know whether there is a slicker way, but yes, that's what I meant. I agree about the assumption of positivity (I assumed it). –  Davide Giraudo Nov 22 '12 at 23:00
    
Okay, thanks for the clarification. –  commenter Nov 22 '12 at 23:26

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