Exercise 1.3.15 (Translation invariance of Lebesgue integral)

If we manage to show that the area under the curve of $f$ and the area under the curve of $f^{\prime }$ are translates of each other

then the assertion will follow by combining the translation invariance of Lebesgue measure and the fact that Lebesgue integral of $f$ and $f^{\prime }$ is equal to the Lebesgue measure of the area under their curves.

• Pick an arbitrary $(z,t)$ from the left hand side. Then, there exists $x\in {ℝ}^d$ such that $z=x-y$ and $t\le f(x)$; in other words $(z,t)\le (x-y,f(x))$. The fact that $(z,t)$ is contained in the right-hand set then follows by definition.
• Pick some $(x,t)$ from the right-hand side. Then $t\le f(x+y)$. In other words, $t\le f(z)$ for $z-y$; thus, $(x,t)+(y,0)$ is contained in $\left\{(x+y,t)\in {ℝ}^{d+1}:0\le t\le f(x+y)\right\}=\left\{(x,t)\in {ℝ}^{d+1}:0\le t\le f(x)\right\}$.

Combining Exercise 1.2.20 and Exercise 1.3.13 yields the desired result.