Lebesgue measure of a triangle

Question

"Proof that the Lebesgue measure of a triangle is the euclidean area."

Well, I know firstly that a triangle is measurable with finite measure, because is closed and bounded, and that the Lebesgue measure is invariant by translations. But I don't know how to prove it basing on what I've just said.

Can someone give me a hand on this problem?

Thanks in advance.

Answer 1

Hint: I presume you know the measure of a rectangle is its area. Two congruent triangles form a parallelogram, and a parallelogram can be dissected and the pieces rearranged to form a rectangle.

EDIT: Another way to do it (assuming you know the measure of a square is its area): bound above and below the number of nonoverlapping squares of side $\epsilon$ that can fit inside the triangle.

Answer 2

I won't work out the full details, but I think the idea is this: the measure of the unit square $S$ with the canonical basis as sides in the plane has Lebesgue measure $\mu(S)=1$ (this should be obvious from the definition of $\mu$). Now consider the linear transformation $\phi$ that takes the canonical basis to two adjacent sides of the triangle $T$ (translate $T$ if needed). Then $\phi(S)$ is a parallelogram of area twice that of $T$, so $2\mu(T)=\mu(\phi(S))=|det\,\phi|\mu(S)=|det\,\phi|$ so it reduces to a linear algebra problem (I'm assuming you have seen why the det comes into play there, otherwise take a look at Folland theorem 2.44 for example)