Value for Jordan Measure of solid triangle In Terrence Tao's 'Introduction to Measure Theory' p.10-11 he poses the following question:
Let $A,B,C$ be three points in $\mathbb{R}^2$
i) Show that the solid triangle with vertices A, B, C is Jordan Measurable
ii) Show that the Jordan measure of the solid triangle is equal to $\frac{1}{2} (\lvert{(B-A)\land(C-A)}\rvert)$ where $\lvert (a,b)\land(c,d)\rvert := \lvert ad-bc \rvert$
I have been able to show the first statement by a bit of a contrived argument using the fact that the union of 2 Jordan measurable sets is Jordan measurable and that the area under a graph is Jordan measurable.
The proof for the second statement seems beyond me in light of Jordan measure and the text.
Any help would be much appreciated!  
 A: Question (2) is very different to question (1), as we need to actually calculate a measure, not just prove that one exists, and (1) and the previously question are useless for this. The only useful tools we have available at this point in the book are:


*

*The Jordan measure of a box  (the product of its intervals).

*Finite additivity (of disjoint measureable sets).

*Translation invariance.


There are no other statements that equate the measure of something (inequalities are useless here), so any proof needs to be rooted in these 3 properties, or you must revert back to the definition of Jordan measure and deal with supremums and infimums. I don't think other answers here understand how little we are actually able to use. 
There are two options. Either 


*

*follow @user400448 and derive integration of triangles from first principles, utilizing the definition of Jordan measure.

*Try to utilize the three statements above, somewhat to an extreme. This is what I describe below.


A right-angle triangle along an axis
Like Tao hints, start with calculating the measure of right angle triangle with one side on an axis. You can try work with the Jordan measure directly here. I tried to avoid this and instead I used symmetry. I first proved these 2 lemmas:


*

*Lemma 1. Reflecting a set over the x-axis preserves the measure of the set.

*Lemma 2. Reflecting a set over the y-axis preserves the measure of the set.


These lemmas are relatively straightforward due to how boxes align with the axes. I'm not including the proof here.
Then, you can show that a box can be split in half by two disjoint right-angle triangles with equal measure. The point of this is that you have managed to measure any right-angle triangle which has a side parallel to an axis. The image below tries to show the steps of proving these lemmas. I'm also leaving out the specific steps here.

In summary, you should arrive at the following lemma:
Let $A = \{(x, y) \in \mathbb{R}^2 : ya \le xb \text{ and } x \le a \}$ be a subset of $\mathbb{R}^2$. In other words A is the set representing the right angle triangle at the origin with a side along the x-axis and a point (a,b). To make life easy, denote such a triangle as:
$$ A = Trig(a,b) \tag{1}\label{1}$$
Then, the Jordan measure of A is given by:
$$
m(A) = m(Trig(a, b)) = \frac{1}{2}ab \tag{2}\label{2}$$
An arbitrary triangle
A triangle at the origin has three points (0, 0), (a, b) and (c, d). Without loss of generality, assume that $a \ge c$. There are then three cases: $d > b$, $d = b$ or $d < b$. You need to deal with each of these cases. The lemma for reflection over the x-axis will allow the last case to follow easily from the first. I'll just cover the case where $d > b$. The set representing this triangle can be expressed as:
$$
T = \{(j, k) \in \mathbb{R}^2 : (kc \le jd) \text{ and } (ka \ge jb) \text{ and } \frac{j - c}{a - c} + \frac{j - b}{d - b} \le 1 \}
$$
This corresponds to the intersection of the areas above/below each of the three lines that intersect to make the triangle:

We have most things in place now.
$T \subseteq B$ where $B = (0, a) \times (0, d)$. (why? Any point in T must also be in B, by some tedious algebra.) Subtracting T from B then gives us:
$$B \setminus T = \{(j, k) \in B : (kc > jd) \text{ or } (ka < jb) \text{ or } \frac{j - c}{a - c} + \frac{k - b}{d - b}  > 1 \}$$
This uses the fact that: not (p and q) ⟺ not p or not q.
You can split this set into three separate unions corresponding to $W_1$, $W_2$ and $W_3$. Again, there is some more tedious work to show that these are all disjoint.
$$
\begin{align}
B \setminus T = W_1 \cup W_2 \cup W_3 = \\
&\{(j, k) \in \mathbb{R}^2 : kc > jd \text{ and } y \le d \} \\
\cup &\{(j, k) \in \mathbb{R}^2 : ka < jb \text { and } x \le a \} \\
\cup &\{(j, k) \in \mathbb{R}^2 : \frac{j - c}{a - c} + \frac{k - b}{d - b} \ge 1 \}
\end{align}
$$
With a bit of algebra, each of these sets can be seen to be translated and reflected to equal a right angle triangle with a side on the x-axis. 
We now can start to measure. 
$$
\begin{align}
m(B \setminus T) 
&= m(W_1 \cup W_2 \cup W_3) \\
&= m(W_1) + m(W_2) + m(W_3) \text{ (as each are disjoint)} \\
&= m(Trig(a,b)) + m(Trig(d, c)) + m(Trig((a - c), (d - b))) \tag{3}\label{3}\\
&= \frac{1}{2}ab + \frac{1}{2}dc + \frac{1}{2}(a - c)(d - b) \\
&= \frac{1}{2}(ab + dc + ad - ab - cd + cb) \\
&= \frac{1}{2}(ad + cb)
\end{align}
$$
The step to arrive at (3) is possible due to translation invariance of Jordan measure (given in the book), and reflection invariance over an axis, which you need to prove, as I mentioned above.
From this we see that the measure of $T$ is:
$$
\begin{align}
m(T) &= m(B) - m(B \setminus T) \\
&= ad - \frac{1}{2}(ad + cb) \\
&= \frac{1}{2}(ad - cb)
\end{align}$$
You now need to handle the two cases when $d=b$ and $d<b$. 
An alternative two this 3 part proof would be to introduce some sort of lemma that allows you to relate the measure of a triangle to its enclosing parallelogram. Whatever you choose, I don't think you can use visual reasoning like rotation invariance or visually dividing up a square into triangles without proving these ideas first. 
A: Hint: show that the Jordan measure is just the area of the triangle by expressing the triangle as a boolean combination of "areas under graphs".
A: I did this question by splitting up the problem into different cases.Then, use area under the graph. What helps is to consider the point, say A, which has the smallest "x coordinate". Then you can split into 3 cases. X(B)=X(C), X(B)>X(C) and X(B)

However, this does require the use of integration I guess, which you can do from first princple. Personally, It just looked to messy and to long for such an elemantry question, but it works. 
A: Though this question has been around for quite some time, all answers have negative score, so I'll throw some ideas for that.
If you did the previous exercise, it showed that the measure of the area under the graph of a function is actually the numerical value of the area, approximated by those elementary sets.
Now, think of the triangle as an area between the straight lines that form its edges, and use the properties of the measure for the difference between those sets (since they're disjoint).

If you didn't do the previous exercise, the general idea was that your boxes were vertical n-dimensional bars (just visualize the Riemann integral picture of bars approximating the area under the graph). 
In the triangle case, use a vertical bar from the x-axis to the upper edge(s), and then subtract a vertical bar that only reach the lower edge(s).
A: The first statement can be shown by the area under graph can be approximated by rectangles as did in Reimann integral. 
The second statement can shown below, suppose AC is horizontal to the x-axis
the plot is here
1/2 (|(B−A)∧(C−A)|)=1/2 (|(x3-x1,y3-y1)∧(x2-x1,0)|)=1/2 (the area of the ACDE (green region))=area of the triangle=the Jordan measure of the solid triangle. 
