What's the Lebesgue measure of:
$A_1 = \left \{ (x, y) \in \mathbb{R}^2: 1 < x^2 + y^2 \leq 9 \right \}$
$A_2 = \left \{ (x, y) \in \mathbb{R}^2: x^2 + y^2 \leq 1 \right \}$
I clearly have problems with intuition what a Lebesgue measure is at all and reading the definition doesn't help me a lot, so if you could explain this clearly and possibly provide some materials to help me apprehend this concept, I'd really be thankful.
EDIT:
If Lebesgue measure coincides with the standard area, does that mean that the measure of $A_1$ is $8 \pi$ and the measure of $A_2$ is $\pi$?
The Lebesgue measure of a set is simply its volume (or in the two-dimensional case the surface area)
$A_2$ is a disk with radius $1$, so $\mathcal{L}(A_2)=\pi$
$A_1$ is a disk with radius $3$, but missing an inner centered disk of radius 1, so $\mathcal{L}(A_2)=9\pi-\pi =8 \pi$.
Your two cases were really easy, as one immediately sees the sets shapes. In more complicated cases, you need a more systematical way to calculate the Lebesgue measure. This is done by approximating your sets with boxes (their Lebesgue measure is trivial) and taking the limit of letting this boxes get smaller (results in better approximations).
This approximation can also be done in two consecutive steps (like it is done in the picture above). First taking boxes with fixed width and optimal height, and then later optimizing the width. (This is called two-dimensional integration).
Then $$\mathcal{L}^2(A_2) =\int_{A_2} d\mathcal{L}^2 = \int_{\{0\le x^2+y^2\le 1\}} d\mathcal{L}^2(x,y) \\ =\int_{-1}^1 d\mathcal{L}(x) \int_{-\sqrt{1-x^2}}^\sqrt{1-x^2} d\mathcal{L}(y)\\ =4 \int_0^1 \sqrt{1-x^2} \, d\mathcal{L}(x)=\pi$$