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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](http://en.wikibooks.org/wiki/Measure_Theory/Riesz%27_representation_theorem) 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"2012-11-22
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    @icurays1 What happens if there is not a topological structure on $X$?2012-11-23

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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.2012-11-22
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    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).2012-11-22
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    Okay, thanks for the clarification.2012-11-22