Showing the set of real values for which the pre-image has measure greater than zero is measure zero The question is stated as follows:
Show that if $f: \mathbb{R} \rightarrow \mathbb{R}$ is measurable, then the set $E = \{x \in \mathbb{R} \ | \ m(f^{-1}(x)) > 0 \}$ has measure zero.
This problem seems simple enough at first glance, and I feel like I had a solution. We begin by showing that $E$ is measurable using a definition from Royden. That is, since $E \subset \mathbb{R}$, and we have
\begin{align}
m^*(\mathbb{R}) = m^*(\mathbb{R} \cup E) + m^*(\mathbb{R} \cup E^C) 
\end{align}
it follows that $E$ is a measurable set. Traditionally, if the inverse of $f$ is defined, then f is bijective and for each $y \in \mathbb{R}$, there should be one $x \in \mathbb{R}$ such that $f(x) = y$. But then for each $y \in \mathbb{R}$, $m(f^{-1}(y)) = m(\{x\}) = 0$. From this, it should follow not only that $E$ is empty, but that $m(E) = 0$. 
Here is where I'm not sure. If we just consider $f^{-1}$ to return the pre-image of a point $y \in \mathbb{R}$, and say that $f$ is a constant function defined $f(x) = c$, then $m(f^{-1}(c)) = m(\mathbb{R}) = \infty > 0$. In this case, however, $E = \{c\}$, which has measure zero. With this new interpretation of $f^{-1}$, I'm not exactly sure how to proceed. If the measure of $E$ was not zero, there would necessarily be an interval $(y_1,y_2) \subset E$ for which each point in the interval $y \in (y_1,y_2)$, we have $f^{-1}(y) = (x_1,x_2)$. I'm not sure what implications this might have, if any. Should I work at obtaining a contradiction here, or should I attempt to prove that $m(E) = 0$ directly?
Edit: As was pointed out in the comments, my argument for the measurability of $E$ is insufficient. Further, my conclusion that an interval lived in $E$ is equally false (thanks Cantor)
 A: Hint: Since Lebesgue measure is $\sigma$-finite it follows that $E$ is actually countable.
A: The function $f$ doesn't even have to be measurable for this to be true.  (Of course, if $f$ isn't measurable, then there may exist $x \in \mathbb{R}$ such that $f^{-1}(x)$ is not a measurable set, but that's OK — any such $x$ won't be in $E$ since it's not true that $m\left(f^{-1}(x)\right)\gt 0.)$
If $f$ isn't required to be measurable, we're interpreting $E$ as $\lbrace x\;\vert\;f^{-1}(x)\text{ is measurable and }m(f^{-1}(x))>0\rbrace$ or, equivalently, $\lbrace x\;\vert\;f^{-1}(x)\text{ has positive measure}\rbrace.$
As @DavidC.Ullrich pointed out, $E$ will actually turn out to be countable.
For positive integers $s$ and $t,$ let $E_{s,t} = \lbrace x\;\vert\; m\left([t,t+1) \cap f^{-1}(x) \right) \gt \frac{1}{s} \rbrace . $
Claim: Each $E_{s,t}$ is finite.  In fact, the cardinality of $E_{s,t}$ is less than $s.$
Proof of claim: Suppose $x_1, \dots, x_n$ are $n$ distinct elements of $E_{s,t};$ we'll show that $n \lt s.$  The sets $f^{-1}(x_1), \dots, f^{-1}(x_n)$ are pairwise disjoint, so
$$1 = m\left([t,t+1)\right) \ge \sum_{k=1}^{n}m\left([t,t+1)\cap f^{-1}(x_n)\right) \gt \frac{n}{s},$$  where the last inequality is because each of the $n$ summands is greater than $\frac{1}{s}.$  It follows that $n \lt s.$
Now define $E_t = \lbrace x\;\vert\; m\left([t,t+1) \cap f^{-1}(x) \right) \gt 0 \rbrace .$  A real number $a$ is greater than $0$ iff there is a positive integer $s$ such that $a$ is greater than$\frac{1}{s};$ so $E_t = \cup \lbrace E_{s,t}\;\vert\; s\text{ is a positive integer}\rbrace.$  This is a countable union of finite sets, so each $E_t$ is countable.
Next we observe that if a subset $A$ of $\mathbb{R}$ has positive measure, then, for some integer $t,$ the set $A \cap [t,t+1)$ has positive measure.  After all, each of the sets $A \cap [t,t+1)$ is the intersection of two measurable sets, so is measurable; but if they all had measure $0,$ then $A$ would be the countable union of measure $0$ sets, so $A$ would have measure $0$ itself.
It follows that $E \subseteq \cup \lbrace E_t\;\vert\;t\text{ is an integer}\rbrace.$  The union here is a countable union of countable sets, so it must be countable.  Its subset $E$ must therefore be countable also. Any countable set has measure $0,$ so we're done.
Finally, a note on the axiom of choice: this argument doesn't use the axiom of choice although it may at first appear to.  (The general theorem that a countable union of finite sets is countable does require the axiom of choice, as does the theorem that a countable union of countable sets is countable.  But in the particular cases used here, the axiom of choice is not needed.)
The reason that choice isn't required here is that the reals are linearly ordered (by the usual ordering), and this linear ordering provides a specific, definable well-ordering of each of our finite sets $E_{s,t}.$  This in turn gives us a way to define a specific counting of each of the sets $E_t,$ and that's all that's needed to conclude that their union is countable.  
