Dual space of L_1

EDIT - I was wrong, it turned out that all I need was the Reisz Representation theorem.

I am looking for a way to understand how $L^{\infty}$ is a realization of the strong dual of $L^1$, but I am only interested in the case when the measure is Lebesgue measure. Most of the proofs that I've (in Rudin and in Treves: TVS, Distributions and Kernels) seem involve the general case, and introduce complexities (such as appeals to the Radon-Nikodym theorem) that I haven't studied yet. I've looked at this section in Rudin's real and complex, but there is a lot of measure theory (I only know the very basic stuff) that I would have to learn before I can understand it. However, since I am only interested in one particular case of $L^1$ (Lebesgue integration) I thought it might be possible that there was a simpler proof out there (especially since, after glancing at the Radon-Nikodym theorem, it seems trivially true if the measure is just the Lebesgue measure throughout - but I'm not really sure if I'm thinking about it in the right way). If any body could direct me towards a resource where I could find such a proof I would be very happy.

This question comes out of trying to understand what Rudin means (in his Functional Analysis) when he shows that a distribution is locally the derivative of a continuous function (6.26): he extends a bounded linear functional $\Lambda_1$ on a subspace of $L^1(K)$ (K compact) by the Hahn-Banach theorem, and then says there there must be some bounded and measurable function g on K so that $\Lambda_1(\phi) = \int g(x) \phi(x) dx$ $(\phi \in D(K)$). I would like to understand this connection more thoroughly, the rest of the proof I understand (I think). I am assuming that this a reference to the duality relationship I asked about my question about above. If I am wrong, then understanding this move in the proof is more important to me than understanding the duality relationship (though that is also interesting to me).