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Apostol: Area of a triangle

  1. May 20, 2006 #1
    Prove that every triangular region is measurable and its area is one half the product of its base and altitude.

    Let Q be a set that can be enclosed between two step regions S and T , so that [tex] S \subseteq Q \subseteq T [/tex] If there is one number c such that [tex] a(S) \leq c \leq a(T) [/tex], then Q is measurable, and [tex] a(Q) = c [/tex]. So we know that [tex] c = \frac{1}{2}bh [/tex]. So there has to be two step regions so that the area of the triangle is between them. We know that every rectangle is measurable, and the [tex] a(R) = bh [/tex]. So how do I prove that [tex] c = \frac{1}{2}bh [/tex]?
  2. jcsd
  3. May 20, 2006 #2


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    I believe you know that if S and T are measurable, then so is S u T and moreover, the formula a(S u T) = a(S) + a(T) - a(ST) holds. Do you also know that if S and T are measurable with S a subset of T, then a(S) < a(T)? That property is refered to as monotonicity.

    Let Q be the triangle. Define:

    [tex]\mathcal{T} = \{T \mbox{ measurable } | Q \subseteq T\}[/tex]

    [tex]\mathcal{T}_R = \{T \mbox{ a union of rectangles } | Q \subseteq T\}[/tex]

    [tex]\tau = \inf \{a(T) | T \in \mathcal{T}\}[/tex]

    [tex]\tau _R = \inf \{a(T) | T \in \mathcal{T}_R\}[/tex]

    [tex]\mathcal{S} = \{S \mbox{ measurable } | S \subseteq Q\}[/tex]

    [tex]\mathcal{S}_R = \{S \mbox{ a union of rectangles } | S \subseteq Q\}[/tex]

    [tex]\sigma = \sup \{a(S) | S \in \mathcal{S}\}[/tex]

    [tex]\sigma _R = \sup \{a(S) | S \in \mathcal{S}_R\}[/tex]

    [tex]c = \frac{bh}{2}[/tex]

    Prove that [itex]\mathcal{T}_R \subset \mathcal{T}[/itex], and from that get [itex]\tau _R \geq \tau[/itex]. Prove [itex]\mathcal{S}_R \subset \mathcal{S}[/itex], and from that get [itex]\sigma _R \leq \sigma[/itex]. Prove that you can cover the triangle with rectangles such that the total area of those rectangles is arbitrarily close to c, and infer that [itex]c \geq \tau _R[/itex]. Likewise, prove that [itex]c \leq \sigma _R[/itex]. Use monotonicity to prove that [itex]\tau \geq \sigma[/itex]. Putting together some of these inequalities gives:

    [tex]c \geq \tau \geq \sigma \geq c[/tex]


    [tex]\sigma = c = \tau[/tex]

    This is exactly what you want to prove.
    Last edited: May 20, 2006
  4. Jul 19, 2010 #3
    Sorry to bump a 4 year old post, but I did not want to start my own thread on the same question. I am working through the Apostol Calculus book now, and I would like to ask some questions on parts of this proof.

    1. Is it sufficient to say that based on Axiom 5 in Apostol's Calculus (Every rectangle R is in M {the measurable set}) [itex]
    \mathcal{T}_R \subset \mathcal{T}

    2. If [itex]
    \mathcal{T}_R \subset \mathcal{T}
    [/itex], how can [itex] \tau _R [/itex] be greater than [itex]\tau[/itex]? How do I prove this?

    Basically, this proof is showing the infimum of the area of [itex]\tau[/itex] is equal to the supremum of the area of [itex]\mathcal{S}[/itex], which is equal to [itex]\matchcal{c}[/itex], correct?

    Also, I don't think I am grasping this:

    \mathcal{T} = \{T \mbox{ measurable } | Q \subseteq T\}
    \mathcal{T}_R = \{T \mbox{ a union of rectangles } | Q \subseteq T\}

    Are both [tex] \matcalc{T} [/tex]s the same? Wouldn't it be better to use a different variable for the measurable set and the union of rectangles set? Again, I don't think I am grasping this properly.
    Last edited: Jul 19, 2010
  5. Sep 4, 2012 #4
    Even though it is a very old topic, I am working through apostol's calculus and got stuck on the same problem.

    What I don't get is how do we do that

    I thought that if we cover Q with a union of T rectangles such that [itex]Q \subseteq T[/itex], the infimum of [itex]\tau_R[/itex] will be bigger than c, because the areas of rectangles we are covering the triangle are at least as big as the area of triangle itself? Please help to wrap my head around this problem :)
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