Proving a set has Lebesgue measure zero

Question

Let $S=\{(x,x):0\le x<1\}\subset\mathbb{R}^2$. Prove $\lambda_2(S)=0$ where $\lambda_2$ is Lebesgue measure on $\mathbb{R}^2$.

Attempt: I want to wrestle with the definition of Lebesgue measure and hopefully will get the result from that. By definition, $\lambda_2(S)=\inf\{\sum_i (b_i-a_i):S\subset\bigcup_i (a_i,b_i]\}$. So in our case, can we simply say $S\subset (x-\epsilon,x+\epsilon)\times (x-\epsilon,x+\epsilon)$ and then $\lambda_2(S)\le (2\epsilon)^2\to 0$? Since the measure of $(x-\epsilon,x+\epsilon)$ is $2\epsilon$.

An alternative idea - we know that the graph of a continuous function on $\mathbb{R}^2$ has measure zero, so I suppose we could argue that $f(x)=x$ in this case would yield a continuous function's graph, hence of measure zero. Any comments would be appreciated!

Answer 1

This seems to be a textbook exercise, so I'll just give you an hint (which is related to what geetha290krm wrote in the comments).

You can rewrite $\lambda_2(S)$ as an integral, namely: $$ \lambda_2(S) = \int_S d\lambda_2(x,y) $$ (This should be easy to see: $f(x,y) = 1$ is a simple function, which allows you to derive an explicit formula of the integral w.r.t. the measure.)

But given the definition of $S$, it is easy to rewrite this formula as a double integral in $x$ and $y$. What can you say about this double integral?

Answer 2

Let $T:\mathbb{R}^2\to\mathbb{R}^2$ denote rotation by $\frac{\pi}{4}$ counter-clockwise. Then

$$S=T(\{0\}\times[0,\sqrt{2})).$$

As $T$ is an isometry, and $\lambda_2$ is invariant under isometries, it follows that

$$\lambda_2(S)=\lambda_2(\{0\}\times[0,\sqrt{2}))=0\cdot\sqrt{2}=0.$$