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I want to show that every self-adjoint operator is unitarily equivalent to multiplication by some $\lambda \in L^{\infty}(X)$ on an $L^2(X)$ for some measure space $X$, $\sigma$ finite.

I've been working with the Borel functional calculus for a while and have made no progress.

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    You are interested in the unbounded case, right? – 2017-02-18
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    @s.harp We only consider bounded operators. – 2017-02-18
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    This is the Spectral Theorem for Bounded [Self-adjoint operators](http://www.unc.edu/math/Faculty/met/specthm.pdf). – 2017-02-18

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For a multiplication operator $T=M_f$ one has a natural functional calculus via $g(T)=M_{g\circ f}$. Hence the usual strategy to obtain the Borel function calculus for a normal, bounded operator is to represent it as a multiplication operator, apply the functional calculus as above and then transform back to the original Hilbert space. Thus I would suggest that you take another look into your material on the Borel functional calculus.