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Are the following true? If so, how does one prove them?

Suppose $f:\Bbb R\to\Bbb R$ is continuous.

(i) $\partial \{f>t\} \subset \{f=t\}$, where $\{f>t\}:=\{x \in\Bbb R:f(x)>t\}$ and $\partial$ denotes the boundary of the set.

(ii) The set $\{f=t\}$ has Lebesgue measure $0$ except for at most countably many $t$.

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up vote 3 down vote accepted

The set $S:=\{f>t\}$ is open, so $\partial S=\overline S\setminus S$. If $x\in \overline S\setminus S$ then $f(x)\leq t$ and $x=\lim_n x_n$, where $x_n\in S$ hence $f(x)\geq t$ so $f(x)=t$.

Let $S_{t,n}:=\{x,f(x)=t\}\cap [n,n+1)$ for $t\in\Bbb R$ and $n\in\Bbb Z$. For $n$ and $k$ fixed, the set $$I_{n,k}:=\{t\mid m(S_{t,n})\geq k^{—1}\}$$ is finite, as the sets $\{S_{t,n}\}$ are pairwise disjoint and contained in $[n,n+1)$. Then $\{f=t\}$ except for $t$ in the countable set $\bigcup_{k\in\Bbb N^*,n\in\Bbb Z}I_{n,k}$.

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Why is $I_{n,k}$ finite? I don't follow your second proof. Could you provide some more details? – Cantor Oct 7 '12 at 20:27
If not, we would have infinitely many subsets of $[n,n+1)$ of measure $k^{—1}$ which are pairwise distinct. – Davide Giraudo Oct 7 '12 at 20:28
Where was continuity used in the second proof? – Cantor Oct 7 '12 at 21:12
It seems nowhere, and that it's true for any map. – Davide Giraudo Oct 7 '12 at 21:13

If there are uncountably many values of $t$ for which the set $\{x : f(x) = t\}$ has positive measure, then the real line has uncountably many pairwise disjoint subsets of positive measure. Consider their intersections with the sets $[n,n+1)$ for $n\in\mathbb Z$. Can it be that for each $n$, only countably many of them have an intersection with that interval that has positive measure? But countably many countable sets add up to countably many sets, so that can't happen. So within $[n,n+1)$ you've got a set $S$ of uncountably many pairwise disjoint sets of positive measure. Within any uncountable set of positive numbers, you can find a countable subset whose sum is more than $1$. But that can't happen because the measure of $[n,n+1)$ is $1$.

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