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In section 1.18 ("The area of an ordinate set expressed as an integral"), Apostol proves two theorems. the first, theorem 1.10, deals with the area of a function's ordinate set; the second, theorem 1.11, deals with the area of the graph of the function of theorem 1.10. (I have attached two excrepts from Apostol's book, one per theorem.)

I am having problems understansing Apostol's logic in theorem 1.11 where he states:

"The argument used to prove Theorem 1.10 also shows that Q’ is measurable and that a(Q’) = a(Q)."

I don see how he could argue this, being that ##Q'=\{(x,y)|a \le x \le b, 0 \le y < f(x) \}## which implies that ##S \subseteq Q'## is not true for all step regions S (since S may contain a point of the graph of ##f(x)## which Q', by definition, can't).

Thanks in advanced for any help.

I am having problems understansing Apostol's logic in theorem 1.11 where he states:

"The argument used to prove Theorem 1.10 also shows that Q’ is measurable and that a(Q’) = a(Q)."

I don see how he could argue this, being that ##Q'=\{(x,y)|a \le x \le b, 0 \le y < f(x) \}## which implies that ##S \subseteq Q'## is not true for all step regions S (since S may contain a point of the graph of ##f(x)## which Q', by definition, can't).

Thanks in advanced for any help.