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I don't know how much about areas you want to prove, and how developed a background the audience is supposed to have, but here is a definition of area of a bounded planar region.


$$\eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D). N_\ve(D)$$

exists. If this is the case, then we define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$.

The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann integrable functions and $D(U,L)$ is the region

$$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace,$$

then $D(U,L)$ is measurable

$$\eA(D(U,L))=\int_a^b \bigl(\, bigl(\; U(x)-L(x)\;\bigr) dx.$$

The next thing to prove is a weak form of the inclusion exclusion inclusion-exclusion principle: if $D_1$, $D_2$ are measurable regions that intersect along the grapf of a $C^1$-function, then $D_1\cup D_2$ is measurable and

$$\eA(D_1\cup D_2)=\eA_(D_1)+\eA(D_2)D_2)=\eA(D_1)+\eA(D_2).$$

1

I don't know how much about areas you want to prove, and how developed a background the audience is supposed to have, but here is a definition of area of a bounded planar region.


$$\eA(D)= \lim_{\ve\to0}\ve^2 N_\ve(D).$$

exists. If this is the case we define the area to be the limit $\eA(D)$, and we set $\eA_\ve(D)=\ve^2N_\ve(D)$.

The first step is to prove $\newcommand{\bR}{\mathbb{R}}$ that if $L,U: [a, b]\to \bR$ are Riemann integrable functions and $D(U,L)$ is the region

$$D_f= \bigl\lbrace\; (x,y)\in [a,b]\times \bR;\;\;L(x) \leq y\leq U(x)\;\bigr\rbrace,$$

then $D(U,L)$ is measurable

$$\eA(D(U,L))=\int_a^b \bigl(\, U(x)-L(x)\;\bigr) dx.$$

The next thing to prove is a weak form of the inclusion exclusion principle: if $D_1$, $D_2$ are measurable regions that intersect along the grapf of a $C^1$-function, then $D_1\cup D_2$ is measurable and

$$\eA(D_1\cup D_2)=\eA_(D_1)+\eA(D_2).$$