The problem states:
Find the measure of the set $\Gamma_f = \{(x, f(x)) \in \mathbb{R}^2 | x \in (0, 1]\}$, for $f : (0, 1] \rightarrow \mathbb{R}$, $f(x) = \frac1x$.
There was a lemma which stated that if $f : [a, b] \rightarrow \mathbb{R}$ is continuous, then $m(\Gamma_f) = 0$ (with help of uniform continuity of the function $f$ on a compact segment). It is not hard to prove similar claim for a continuous function $f : \mathbb{R} \rightarrow \mathbb{R}$ by writing $\mathbb{R}$ as the union of segments $[n, n + 1]$, where $n$ are integers.
However, the given function $f$ is not defined on a compact set, and when I tried defining $g(0) = f(0) = \infty$ and $g(x) = f(x)$, otherwise (to keep the continuity with an idea that $\Gamma_f \subseteq \Gamma_g$), the function $g$ was not uniformly continous on $[0,1]$ (unless I am mistaken).
Then I've spent some time still trying to prove that $m(\Gamma_f) = 0$ by covering the set $\Gamma_f$ with a set whose measure is $0$.
Finally, I've noticed that: $$\Gamma_f = \sqcup^{\infty}_{n = 1}(\frac1{n+1}, \frac1n]\times[n, n + 1).$$ Then, $$m(\Gamma_f) = m(\sqcup^{\infty}_{n = 1}(\frac1{n+1}, \frac1n]\times[n, n + 1)) = \sum^{\infty}_{n = 1}m((\frac1{n+1}, \frac1n]\times[n, n + 1))) = \sum^{\infty}_{n = 1}m((\frac1{n+1}, \frac1n]) m([n, n + 1))) = \sum^{\infty}_{n = 1}\frac1{n(n + 1)} = 1.$$
So, is the previous union indeed disjoint, is it equal to $\Gamma_f$ and (if I've made a mistake in this post) is it true that in $\mathbb{R}^2$ every continuous function has measure of its graph equal to zero?