Lebesgue measure for the graph of a continuous function $f \colon A \subseteq \mathbb{R}^n \to \mathbb{R}^m$

Question

My guess is that I must prove that $A \times \mathbb{R}^m$ has measure zero in $\mathbb{R}^n \times \mathbb{R}^m$.

Now, knowing that f is continuous on A, whatever succesion $\{x_n\} \subset \mathbb{R}^n$ s.t. $\{x_n\} \to x \implies f(\{x_n\}) \to f(x)$ (i.e.: $f(x_n) \subset \mathbb{R}^m$).

Can we take $C_n = \{(x_n,f(x_n)\} \subset \mathbb{R}^n\times \mathbb{R}^m$ as a cover for $A \times \mathbb{R}^m$? Can I prove that $\sum vol(C_i) < \varepsilon$?

Thank you

Answer 1

Be careful! The graph of $f$ is not $A \times \mathbb{R}^m$ in general. You might be saying that the graph is contained in this set, so if we can show $A \times \mathbb{R}^m$ is null, then the graph must be too, but this might be false! As a cute exercise, you might try to show this, but I'll leave a big hint below the fold:

Consider any $f : [0,1] \to \mathbb{R}^2$. Then you're suggesting we consider the set $[0,1] \times \mathbb{R}^2 \subseteq \mathbb{R}^3$. Can you show this set has infinite measure?

As is mentioned in the comments, we also can't rely on the image of $f$ being small! After all, there are continuous injective functions from $\mathbb{R}^1 \to \mathbb{R}^2$ whose images have positive measure! See here, for instance.

So then we'll have to use some special property of the graph. One that comes to mind is that it's "sparse" in the sense that, for a fixed $x \in A$, there is exactly one $y$ so that $(a,y)$ is in the graph. We can leverage this, and let's see how:

---

Let $\Gamma = \{ (a,f(a)) \mid a \in A \}$ be the graph of $f : A \to \mathbb{R}^m$ for $A \subseteq \mathbb{R}^n$. Moreover, let $\chi_\Gamma$ be the characteristic function of $\Gamma$.

Then if $\mu^d$ is the $d$-dimensional lebesgue measure, we have:

$$ \begin{align} \mu^{m+n} \Gamma &= \int \chi_\Gamma \ d \mu^{m+n} \\ &= \int \chi_\Gamma \ d (\mu^m \times \mu^n) \\ &\overset{(1)}{=} \int \left ( \int \chi_\Gamma(x,y) \ d \mu^n(y) \right )\ d \mu^m(x) \\ &\overset{(2)}{=} \int \mu^n(\{f(x)\}) \ d\mu^m(x) \\ &= \int 0 \ d \mu^m(x) \\ &= 0 \end{align} $$

where in step $(1)$ we've applied (the unsigned version of) Fubini's Theorem, and in step $(2)$ we've evaluated the inner integral. We're thinking of $\chi_\Gamma(x,y)$ as a function of $y$ with fixed $x$. But this is where we use the "sparseness" I alluded to earlier! For fixed $x$ there's exactly one choice of $y$ for which $\chi_\Gamma(x,y) \neq 0$: namely $y = f(x)$. So this integral evaluates to the measure of $\{ f(x) \}$, which is (of course) $0$.

---

I hope this helps ^_^