Reed & Simon give the following version of the spectral theorem (Theorem VII.3):
Let $A$ be a bounded self-adjoint operator on $H$, a separable Hilbert space. Then, there exist measures $\{\mu_n\}_{n=1}^N$ ($N = 1, 2, \ldots$ or $\infty$) on $\sigma(A)$ and a unitary operator $$U: H \rightarrow \bigoplus_{n=1}^N L^2(\mathbb{R}, d\mu_n)$$ so that $$(UAU^{-1} \varphi)_n(\lambda) = \lambda \varphi_n(\lambda)$$ where we write an element $\varphi \in \bigoplus_{n=1}^N L^2(\mathbb{R}, d\mu_n)$ as an $N$-tuple $$(\varphi_1(\lambda), \ldots, \varphi_N(\lambda)).$$ This realization of $A$ is called the spectral representation.
According to this theorem, every bounded self-adjoint operator is unitary equivalent to the map $$\varphi(x) \mapsto x\varphi(x).$$
However, in other versions of the theorem I have come across (for example, on Wikipedia) the multiplication operator is more general than multiplication by the identity function. Instead, we have that $A$ is unitarily equivalent to an operator $T_f$ whose action is $$\varphi(x) \mapsto f(x)\varphi(x).$$
Which is the correct form? Are they equivalent?