Converse of f measurable $\Rightarrow \forall a, ${f=a} is measurable is false? When is {f=a} measurable but f is not? Is this one of the times that including infinity is problematic? I'm trying to understand the definition of measurable function.
 A: In fact there are nonmeasurable one-to-one functions.  Since singletons and the empty set are measurable, $\{x: f(x)=a\}$ is always measurable.
For example, let $\{x_\alpha: \alpha \in A\}$ be a Hamel basis of $\mathbb R$ over the rationals $\mathbb Q$, and let $f$ be the function that is linear over $\mathbb Q$, interchanges two basis elements $x_1$ and $x_2$, and leaves the other basis elements fixed.
This is easily seen to be one-to-one.  For convenience we may choose $x_1$ and $x_2$ so that $x_1 - x_2$ is rational.  Then $\{x: f(x) = x\}$ is a Vitali set (i.e. a set that contains
exactly one element from each equivalence class of ${\mathbb R}/\mathbb Q$), and therefore is not measurable.
A: One counterexample would be an injective function $f$, for which $\{x: f(x)=a\}$ is definitely measurable for all $a$, being either a singleton or empty, but which is nevertheless not a measurable function.
If you don't care what $\sigma$-algebra you're working with, you could define $f:\mathbb{R}\to\mathbb{R}: x\mapsto x$, where the domain $\mathbb R$ has the $\sigma$-algebra of countable or cocountable sets, and the codomain's $\sigma$-algebra is the usual one.  On the other hand, it's still possible to produce a counterexample when the domain has the usual $\sigma$-algebra too. 
First, we construct a partition of the interval $[0,1)$ into countably many nonmeasurable sets.  (This construction is found in Folland's "Real Analysis", Section 1.1.)  Let two points in $[0,1)$ be equivalent if their difference is rational; choose a representative from each equivalence class and call the set of them $N$.  Then for each rational number $r$ in $[0,1)$, denote by $N_r$ the set $\{x+r\pmod 1: x\in N\}$.  These $N_r$ are all disjoint (if not we'd have two elements of $N$ differing by a rational number), their union is $[0,1)$ (because each $x\in[0,1)$ differs by a rational number from some element of $N$), and they are nonmeasurable (because, if any were, they all would have the same measure, and we'd have $1 = m([0,1)) = \sum_r m(N_r) = \sum_r m(N)$, which would be $0$ or $\infty$ according as $m(N)=0$ or $m(N)>0$).
To construct $f:[0,1)\to\mathbb{R}$, pick your favorite bijection $h:\mathbb{Q}\cap[0,1)\to \mathbb{Z}$, and define $f$ by 
$$f(x) = x + h(r),\text{ where $r\in\mathbb{Q}\cap[0,1)$ is such that $x\in N_r$.}$$
This is injective, because $f(x) \equiv x\pmod{1}$, and hence all its fibers are measurable sets, but $f^{-1}([0,1)) = N$ is not measurable, so $f$ cannot be a measurable function.
