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If $A$ is a rectangle in $\mathbb{R}^n$ and if we let $f$ be continuous, then how can we show that the graph of $f$ has measure zero in $\mathbb{R}^{n+1}$?

We may define that $A$ is a subset of $\mathbb{R}^n$ and the graph of $f: A\to \mathbb R$ is the set given $\mbox{graph}(f) := \{(x,y) \in \mathbb{R}^{n+1} : f(x) = y\}$.

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2 Answers 2

up vote 6 down vote accepted

First assume that $A$ is compact; then $f$ is uniformly continuous on $A$.

Hence fix an $\epsilon$ and pick a $\epsilon_1$ to be decided later so that for some $\epsilon_2$, we have that any $|x-y| < \delta$ implies that $|f(x)-f(y)|< \epsilon$.

Now, note that the measure of the graph of $f$, denoted by $|\Gamma(f)|$, has bound $$|\Gamma(f)| \leq 2 \epsilon |B(0, \delta)| N(\delta)$$ Where $N(\delta)$ denotes the number of balls with radius $\delta$ it takes to cover $A$ and $|B(0,\delta)|$ is the measure of the ball of radius $\delta$ in n dimensions.

Recall that $|B(0,\delta)| \leq C \delta^n$. Also, if $A$ has side lengths $l_i$ in dimension $i$, then $$N(\delta) \leq C \prod_{i=1}^n \frac{l_i}{\delta}$$ (I threw in the constant because I may have been a little sloppy with that bound) Thus $$|\Gamma(f)| \leq 2 K \epsilon$$ for some constant $K$. But $\epsilon$ was arbitrary, hence the result.

For general $A$, since $|\Gamma(f)| = 0$ on every compact $A_n$, $|\Gamma(f)| = 0$.

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We can write $A$ as a countable union of sets of the form $A_k:=[a_1,a_1+1]\times \ldots \times [a_n,a_n+1]$ where $a_i$ are integers.

We can write $A_k$ as an union of $N^n$ cubes of square $N^{-1}$. We denote them $C_{k,j}$. Then the graph of $f$ restricted to $A_k$ is covered by $C_{k,j}\times 2\omega(N^{-1})$, where $\omega(\cdot)$ is the modulus of continuity of $f$. Then $m((A_k\times \mathbb R)\cap \mathrm{Gr}(f))\leq N^n2\omega(N^{-1})N^{-n}=2\omega(N^{-1})$ so $m((A_k\times \mathbb R)\cap \mathrm{Gr}(f))=0$ for each $k$ and $m(\mathrm{Gr}(f))=0$.

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