Show that $\lambda^2([0,1]\times[0,1])=1$ where $\lambda^2$ is the two-dimensional Lebesgue measure

Question

I need to show that $\lambda^2([0,1]\times[0,1])=1$ where $\lambda^2$ is the two dimensional Lebesgue measure using only that $\lambda^2$ is a measure and that $\lambda^2([a_1,b_1)\times[a_2,b_2))=(b_1-a_1)(b_2-a_2)$.

I have tried considering $[0,1]\times[0,1] = \bigcap^\infty_{n=1} [0,1+1/n) \times[0,1+1/n)$, but am unable to see how to get any further.

Answer 1

Just straight up use your formula :

$$\lambda^2 ([a_1, b_1)\times [a_2, b_2)) = (a_1 - b_1)(a_2 - b_2) $$

With $a_i = 0$ and $b_i = 1 + \epsilon $. If you take the limit of $\epsilon \rightarrow 0$ you get

$$\lambda^2 ([0, 1]\times[0, 1]) = (1 - 0)(1 -0) = 1 $$

Answer 2

From $[0,1] \times [0,1] \subset [0,1+1/n) \times [0,1+1/n)$ you get $$\lambda^2( [0,1] \times [0,1] ) \le (1+1/n)^2.$$ On the other hand, $[0,1] \times [0,1] \supset [0,1-1/n) \times [0,1-1/n)$ yields $$\lambda^2( [0,1] \times [0,1] ) \ge (1-1/n)^2.$$