In the definition of measurable function in Rudin's book, he defines measurable function from a measurable space $X$ to a topological space $Y$ as the inverse image of every open set in the range space is measurable in the domain space, the definition is equivalent to the common definition of measurable function when $Y$ is equipped with borel set ( the smallest $\sigma-$ algebra containing open sets). However, if the $\sigma-$ algebra containing open sets on $Y$ is bigger than Borel set, then this definition gives a broader range of measurable function, can anyone tell me why Rudin defines like that?
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While no one here can speak on Rudin's behalf, it is probably safe to assume that he did so to keep the material as simple as possible, since his books are mainly concerned with functions mapping to $\mathbb{R}$ or $\mathbb{C}$.
The usual definition of measurable function is:
Definition: Let $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ be measurable spaces. A function from $f\colon X\rightarrow Y$ is measurable if for all $E$ in $\mathcal{N}$, $f^{-1}(E)$ is in $\mathcal{M}$.
...and here is a statement (and proof) that gets you from Rudin's definition to the one above (assuming $\mathcal{N}$ is the Borel sets):
Theorem: Let $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ be given measure spaces. Assume $\mathcal{N}$ is generated by $\mathcal{E}$ (that is to say, $\mathcal{N}=\sigma(\mathcal{E})$ is the smallest $\sigma$-algebra containing $\mathcal{E}$). Let $f\colon X\rightarrow Y$ be a function. If for every $E$ in $\mathcal{E}$, $f^{-1}(E)$ is in $\mathcal{M}$, $f$ is measurable.
Proof: Consider the set $$ \mathcal{A}=\left\{ E\subset Y\colon f^{-1}(E)\text{ is in }\mathcal{M}\right\} . $$ Check that $\mathcal{A}$ is stable under complements and countable unions to conclude that it is a $\sigma$-algebra. By the assumption, $\mathcal{E}\subset\mathcal{A}$. Since $\mathcal{E}$ generates $\mathcal{N}$, $\mathcal{E}\subset\mathcal{N}\subset\mathcal{A}$, as desired.
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