Lebesgue measure of graph of $\sin{\frac{1}{x}}$ on $[0,1]$ I am working on something and read that measure of graph of a continuous function on compact sets is zero. Now, I tried to do it for non continuous functions but the set of discontinuities have measure zero. I took $\sin{\frac{1}{x}}$ for $x\neq 0$ and $0$ when $x=0$, My intuition is that it shouldn't be of measure zero since it oscillates very rapidly close to zero. Is it true? How can i prove it?
P.S Please suggest me a reference which deals with measures of graphs. 
 A: It cannot have a measure greater than zero.
Proof: Be $\epsilon>0$ arbitrary. Then the measure of the graph clearly is the sum of the measure of the graph on $[0,\epsilon/2]$ and the measure on the graph on $[\epsilon/2,1]$.
Now on $[\epsilon/2,1]$ the function is continuous, therefore the measure is zero. On the other hand, on $[0,\epsilon/2]$, the graph is a subset of the rectangle $[0,\epsilon/2]\times [-1,1]$, which has measure $\epsilon$. Therefore the measure of that part of the graph is $\le\epsilon$.
Therefore the measure of the graph of $\sin\frac1x$ on $[0,1]$ is $\le\epsilon$ for any $\epsilon>0$. Thus the measure must be $0$
A: For non-negative measurable functions $f: \Bbb{R}^n \to [0, \infty]$ the Lebesgue measure of the set
$$
\mathcal{G}(f) = \{ (x,y) \in \Bbb{R}^n \times [0, \infty) \mid 0 \leq y < f(x)\} \subset \Bbb{R}^{n+1}
$$
is the integral of $f$, so
\begin{equation}
(*) \quad \int_{\Bbb{R}^n} f \, dm = m_{n+1}(\mathcal{G}(f))\,.
\end{equation}
It turns out that if we define a set
$$
\tilde{\mathcal{G}}(f) = \{ (x,y) \in \Bbb{R}^n \times [0, \infty) \mid 0 \leq y \leq f(x)\} \subset \Bbb{R}^{n+1}\,,
$$
then $m_{n+1} (\tilde{\mathcal{G}}(f)) = m_{n+1} (\mathcal{G}(f))$. From this it follows that the set (graph of $f$)
$$
A = \{ (x,f(x)) \in \Bbb{R}^n \times [0, \infty) \mid x \in \Bbb{R}^n \} )
$$
has Lebesgue measure zero. I think this can be generalized to graphs of arbitrary measurable functions.
Proof of $(*)$. First, from the definition of integral
\begin{align}
\int f &= \sup_\varphi \{ \int \varphi \mid \varphi \leq f\}\,,
\end{align}
where the supremum is taken over all simple functions for which $\varphi \leq f$. Then it's easy to show that
\begin{align}
\sup_\varphi \{ \int \varphi \mid \varphi \leq f\} &= \sup_{\varphi \leq f} m_{n+1}(\mathcal{G}(\varphi))\,.
\end{align}
Now let $(\varphi_k)$ be a increasing sequence of simple functions s.t. $\varphi_k \to f$. Then $\mathcal{G}(\varphi_k) \subset \mathcal{G}(\varphi_{k+1})$, and
$$
\mathcal{G}(f) = \bigcup_{k \in \Bbb{N}} \mathcal{G}(\varphi_k)\,.
$$
The set $\mathcal{G}(f)$ is measurable because the sets $\mathcal{G}(\varphi_k)$ are measurable. And from the continuity of $m_{n+1}$ we have 
$$
m_{n+1}(\mathcal{G}(f)) = \lim_{k \to \infty} m_{n+1}(\mathcal{G}(\varphi_k))\,.
$$
Because for every simple function s.t. $\varphi \leq f$ we have $\mathcal{G}(\varphi) \subset \bigcup \mathcal{G}(\varphi_k)$, it follows that
\begin{align}
m_{n+1}(\mathcal{G}(\varphi)) & \leq m_{n+1} \left( \bigcup_{k \in \Bbb{N}} \mathcal{G}(\varphi_k) \right) = \lim_{k \to \infty} m_{n+1}(\mathcal{G}(\varphi_k))\,.
\end{align}
Also
$$
\sup_{\varphi \leq f} m_{n+1}(\mathcal{G}(\varphi)) \geq \lim_{k \to \infty} m_{n+1} (\mathcal{G}(\varphi_k))\,,
$$
so we have established that
$$
\sup_{\varphi \leq f} m_{n+1}(\mathcal{G}(\varphi)) = \lim_{k \to \infty} (\mathcal{G}(\varphi_k)) = m_{n+1}(\mathcal{G}(f))\,,
$$
so $\int f = m_{n+1}(\mathcal{G}(f))$.
To prove that $m_{n+1}(\mathcal{G}) = m_{n+1}(\tilde{\mathcal{G}})$ take $c > 1$ and then it follows that
$$
m_{n+1}(\tilde{\mathcal{G}}(f)) \leq m_{n+1} (  \mathcal{G}(cf)) = c m(\mathcal{G}(f))\,,
$$
and the claim follows by taking $c \to 1$.
