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Prove the given sets are measurable and have zero area

by RoRi

Prove measurability and establish that the area is zero for each of the following.

  1. Any set consisting of a single point.
  2. Any set consisting of finitely many points in the plane.
  3. The finite union of line segments in the plane.
  1. Proof. Since all rectangles are measurable with area equal to hk where h and k are the lengths of the edges of the rectangle, a single point is measurable with area 0, since a point is rectangle with h=k=0. \qquad \blacksquare
  2. Proof. We prove by induction on n, the number of points. For the case n=1, the statement is true by part (a). Now, assume it is true for some n=k \in \mathbb{Z}_{>0}. Then, we have a set S \in \mathcal{M} of k points in the plane and a(S) = 0. Let T be a point in the plane. By part (a), T \in \mathcal{M} and a(T) = 0. Thus,

        \[ S \cup T \in \mathcal{M} \qquad \text{and} \qquad a(S \cup T) = a(S) + a(T) - a(S \cap T). \]

    But, S \cap T \subseteq S, so

        \[ a(S \cap T) \leq a(S) \quad \implies \quad a(S \cap T) \leq 0 \quad \implies \quad a(S \cap T) = 0. \]

    (Where Axiom 1 of area guarantees us that a(S \cap T) cannot be negative.) Thus, a(S \cup T) =0. Hence, the statement is true for k+1 points in a plane, and thus for all n \in \mathbb{Z}_{>0}. \qquad \blacksquare

  3. Proof. We again use induction, now on n, the number of lines in a plane. For n=1, we let S be a set with one line in a plane. Since a line is rectangle and all rectangles are measurable, we have S \in \mathcal{M}. Further, a(S) = 0 since a line is a rectangle with either h= 0 or k =0, and so in either case, hk = 0. Thus, the statement is true for a single line in the plane, the case n=1.
    Assume then that it is true for n=k \in \mathbb{Z}_{>0}. Let S be a set of k lines in the plane. Then by the induction hypothesis, S \in \mathcal{M} and a(S) = 0. Let T be a single line in the plane. By the case n=1 above, T \in \mathcal{M} and a(T) = 0. Thus, S \cup T \in \mathcal{M} and a(S \cup T) = 0 (since a(S) = a(T) = a(S \cap T) = 0). Hence, the statement is true for k+1 lines in a plane, and so for all n \in \mathbb{Z}_{>0}. \qquad \blacksquare

7 comments

  1. DG says:

    @Waleed

    Because the intersection of S and T would be the set of points <> and T. Therefore, S intersection T is a subset of S.

  3. Anonymous says:

    A line segment doesnt necessarily has $h$ = 0 and $k$ = 0. For example, take a line segment on the line x = y. Should we not prove congruence before assuming that?

  4. Jay says:

    The intersection of S and T can either be empty or non-empty. If it’s empty, then it’s a subset of S. If it’s non-empty, it must be that T is contained in S.

  5. Ran says:

    I’m sorry, I don’t understand why the intersection of S and T is a subset of S, in part (b).
    Thanks, by the way, for publishing your solutions!

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