# Prove that, the graph of a measurable function is measurable and has Lebesgue measure zero.

Prove that, the graph of a measurable function is measurable and has Lebesgue measure zero.

I saw some proofs in the internet, if the function is continuous. What is the relation between a continuous function and a measurable function, must they be equal $\mu-a.e.$, or is this approach useless.

Must it have countable discontinuities ? then we could show it like in the continuous case or can you give any hints ?

• Just search here for "graph measurable function". Commented Feb 14, 2014 at 15:38
• You mention Lebesgue measurable only once, but should use it throughout. If $X$, $Y$ are arbitrary measure spaces, then constant functions $x\mapsto c\in Y$ are measureable and the measure of the graph is $\mu_X(X)\cdot \mu_Y(\{c\})$ and possibly $>0$. Commented Feb 14, 2014 at 16:02
• For anyone who sees this in the future: first prove it for bounded functions on a bounded domain, then for bounded functions on an arbitrary domain, and finally for arbitrary (possibly unbounded) measurable functions. Commented Dec 14, 2022 at 11:17

Assume that $$f:\mathbb{R}^d \to \mathbb{R}$$ is measurable. We prove that $$\left \{ (x,f(x)):x\in \mathbb{R}^d \right \}$$ is a Lebesgue null set.

It is enough to prove the claim when $$f:B\to \mathbb{R}$$ is measurable and $$B\subset \mathbb{R}^d$$ is a box (a product of intervals). Then, since $$\mathbb{R}^{d}$$ is a countable union of boxes, the claim will follow.

Let $$G:=\left \{ (x,f(x)):x\in B \right \}$$, and for $$n\in \mathbb{Z}$$, let $$F_n:= G\bigcap (B\times[n,n+1])$$. It is enough to prove that $$F_n$$ is null for every $$n\in \mathbb{Z}$$, so without loss of generality, we show it for $$F:=F_0$$.

Fix some $$k\in \mathbb{N}$$, and let $$I_j:=[\frac{j}{k},\frac{j+1}{k}], j=0,\ldots,k-1$$. Then, $$m(G\cap (B \times I_j))\le m(f^{-1}(I_j)\times I_j)\le \frac{1}{k}m(f^{-1}(I_j))$$ $$\implies m(F)=m\left(\bigcup_j G\cap (B \times I_j) \right)=\sum_j m(G\cap (B \times I_j))\le\frac{1}{k} \sum_j m(f^{-1}(I_j))$$ $$=\frac{1}{k}m\left(\bigcup_j f^{-1}(I_j)\right)=\frac{1}{k}m(f^{-1}([0,1]))\le\frac{1}{k}m(B).$$ Since $$k$$ was arbitrary we are done.

Hint: since you know it for continuous functions, you may find Lusin's theorem useful.

• This was the first thing that i thought. But maybe there is an easier way Commented Feb 14, 2014 at 15:48

Let $F=\{\,(x,f(x))\mid x\in \mathbb R\,\}$. First show that $F\cap([a,b)\times[c,d))$ has measure zero for all $a<b, c<d$. Indeed for $n\in \mathbb N$, $1\le i\le n$, let $$F_i=F\cap\left([a,b)\times[c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\right).$$ Then \begin{align}\mu(F_i)&\le \mu\bigl([c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\bigr)\cdot\mu\bigl(f^{-1}([c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\bigr)\\ &=\frac{\mu\bigl([c,d)\bigr)}{n}\cdot\mu\bigl(f^{-1}([c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\bigr)\end{align} and hence \begin{align}\mu\bigl(F\cap([a,b)\times[c,d))\bigr)&=\sum_{i=1}^n\mu(F_i)\\&\le \frac{\mu\bigl([c,d)\bigr)}n\sum_{i=1}^n\mu\left(f^{-1}([c+\tfrac{(i-1)(d-c)}{n},c+\tfrac{i(d-c)}{n})\right)\\ &=\frac{\mu\bigl([c,d)\bigr)}n\mu\bigl((f^{-1}([c,d))\bigr)\\ &\le \frac1n{\mu\bigl([c,d)\bigr)}\mu\bigl([a,b)\bigr).\end{align} As $n$ is arbitrary and the two intervals have finite measure, we conclude
$$\mu\bigl(F\cap([a,b)\times[c,d))\bigr)=0.$$ We can cover $\mathbb R\times \mathbb R$ with countabley many such rectangles, hence also $\mu(F)=0$.

• You're solution helped me but I think it is not accurate. Note that $\mu (f^{-1}(\cdots))$ may be $+\infty$, so you need to take the measure of its intersection with the interval $[a,b]$. Commented Jan 7, 2015 at 13:29
• Moreover, you should work with $\mu ^*$ rather than with $\mu$ because you still don't know that $F$ is measurable. Commented Jan 7, 2015 at 13:45