Lebesgue-measure in $\mathbb{R}^2$

Question

For $f : [0,1] \to \mathbb {R}_{\ge 0}$ let $$A = \{ (x,y) \in [0,1] \times \mathbb {R}_{\ge 0}\,|\,0 \le y \le f(x) \}$$

and $$B = \{ (x,y) \in [0,1] \times \mathbb {R}_{\ge 0}\,|\,f(x) \le y \le f(x)+1 \} .$$

Let $\lambda$ be the Lebesgue-measure in $\mathbb{R}^2$. Which of these following claims is true for all continuous $f\colon [0,1] \to \mathbb {R}_{\ge 0}$?

a. $\lambda (B) = 1$

b. $\lambda (A) = \int _{[0,1]} f(x)\,\mbox {d}x$

c. $\lambda (A) = 1$

d. $\lambda (B) = \int _{[0,1]} f(x)\,\mbox {d}x$

I think that a and b are true, and that c and d are false. Is that correct?

Answer 1

It is correct. We just use Fubini: $$ \lambda(B) = \int_0^1 \int_{f(x)}^{f(x)+1} 1~\mathrm{d}y ~\mathrm{d}x = \int^1_0 f(x)+1-f(x)~\mathrm{d}x = \int^1_0 1~\mathrm{d}x = 1 $$ $$ \lambda(A) = \int^1_0 \int^{f(x)}_0 1~\mathrm{d}y \mathrm{d}x = \int^1_0 f(x)~\mathrm{d}x $$ By the way: To see that c and d are false consider $f(x) = 2$ and the (correct) formulas we calculated above.