Suppose $f \colon [0,1] \to \mathbb{R}$ is a $B_{[0,1]}/B_{\mathbb{R}}$ measurable function.
- Show that $g \colon [0,1]^2 \to [0, \infty]$ defined as $g (x,y) \colon = |f(x) - f(y)|$, $x,y \in [0,1]$ is measurable with respect to $B_{[0,1]} \otimes B_{[0,1]}/B_{[0,\infty]}$
- Suppose U and V are two t.t.d. Uniform [0,1] random variables defined on some probability space ($\Omega, \mathrm{F}, \mathrm{P}$). Assume that $\mathbb{E}([|f(U) - f(V)|] < \infty$, then show that the random variable $f(U)$ is integrable.
I have done the first part using measure theoretic induction.
The first part can be shown by observing that $[0,1] \rightarrow \mathbb{R}^2$ via $(x,y) \mapsto (f(x),f(y))$ is measurable on product-sets. Since these sets generate the product-$\sigma$-algebra (and her this is equal to the Borel-$\sigma$-algebra on $\mathbb{R}^2$) this map is measurable. As a composition of maps, we see that $(x,y) \mapsto |f(x)-f(y)|$ is measurable, because $(v,w) \mapsto |v-w|$ is measurable as a continuous map.
I suppose that $U$ and $V$ should be i.i.d., that means here independent. This implies that the distribution of $(U,V)$ is just $\lambda^2$ on $[0,1]^2$. So $$\mathbb{E}(|f(U)-f(V)|) = \int_0^1 \int_0^1 |f(x)-f(y)| \, \mathrm{d} x \mathrm{d} y < \infty.$$ Thus, we know by Fubini's theorem that for almost all $y \in [0,1]$ $$\tag{1}\int_0^1 |f(x)-f(y)| \, \mathrm{d} x < \infty.$$ For some fixed $y \in [0,1]$ satisfying (1) we have therefore $$\int_0^1 |f(x)| \, \mathrm{d} x \leq |f(y)| + \int_0^1 |f(x)-f(y)| \, \mathrm{d} x < \infty$$ by the triangle inequality.