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Is it correct that two continuous functions $f,g: \mathbb{R}^n \to X$, where $X$ is a topological space, cannot differ only on a set of measure zero? So as a consequence, for instance, there is at most one continuous representative of an equivalence class in $L^2$?

Eric Auld
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    Is $X$ an arbitrary topological space? – mfl Jun 16 '14 at 23:30
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    If $X$ is an arbitrary topological space then of course you can just give it the indiscrete topology. You want to consider functions to $\mathbb{R}$, for example, and then the statement is true: the point is that their difference cannot be supported on a set of measure zero, and in fact the support of a continuous function $\mathbb{R}^n \to \mathbb{R}$ contains an open set. – Qiaochu Yuan Jun 17 '14 at 00:12

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It's better to put the measure aside for the moment, and consider the following statement, in which $X$ and $Y$ are topological spaces.

If $f:Y\to X$ and $g:Y\to X$ are continuous, and $X$ is a Hausdorff space, then the set $A=\{y: f(y)\ne g(y)\}$ is open in $Y$.

(Note that in $\mathbb R^n$, an open set of zero Lebesgue measure must be empty.)

With measure out of the way, the proof is straightforward. Pick $y\in A$ and let $U$ and $V$ be disjoint neighborhoods of $f(y)$ and $g(y)$, respectively. By continuity, there is a neighborhood $W$ of $y$ such that $f(W)\subset U$ and $g(W)\subset V$. Hence $W\subset A$. $\quad \Box$

As Qiaochu Yuan said, we can't allow $X$ to be an arbitrary topological space. For example, let $X$ be The Line with two origins and let $Y=\mathbb R$. There are two natural continuous maps from $\mathbb R$ to $X$ which agree everywhere except at $0$: one sends $0$ to $p$, the other to $q$.

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