Take $A \subset [0,1]$ that is measurable and $L_1(A)>0$.
Then take $f(x) = L_1(A \cap [0,x])$, for every $x \in [0,1]$
Calculate $\int_A f(x) dL_1(x)$.
*$L_1 = $ Lebesgue measure.
As I understand it:
$$\int_A f(x) dL_1(x) = \int_A \int_{A \cap [0,x]} 1 \ dL_1(x) \ dL_1(x)$$
But I don't understand how to integrate over $A \cap [0,x]$ since it can be not continous (I don't know where A exists and where it doesn't since I only know it's measurable). That problem occurs in both integrals.
I found a problem regarding similar mathematical objects here: $f: [0,1] \to \mathbb{R}$ absolutely continuous, $f' \in \{0,1\}$ (a.e.), $f(0)=0$. Prove that for some measurable subset $A$, $f(x)=m(A \cap (0,x))$. From what I understad, we can assume that every $f$ that is:
- absolutely measurable,
- described over $[0,1]$,
- $f' \in \{0,1 \}$,
- $f(0) = 0$
is equal to: $f(x) = m(A \cap [0,x])$ for our $A$ ($A$ is measurable and $ A \subset [0,1]$) and $x$: ($x \in [0,1]$)
- $A$ is measurable and $ A \subset [0,1]$,
- $x \in [0,1]$.
Maybe in the case stated above it can be argued that $f(x) = m(A \cap [0,x])$ also?
Any help would be appreciated.
There is a convenient trick based on symmetry you can use to evaluate this integral, but I think a more general approach based on the general change of variables theorem can also work, which would probably let you compute other related integrals.
Let $m$ be the Lebesgue measure and define a measure $\mu$ by $\mu(E) = m(A\cap E)$. The problem can be phrased as we want to compute $$ \int_0^1 \mu([0,x])\,d\mu(x). $$ Unwrapping the definition and using Fubini's theorem we observe $$ \int_0^1 \mu([0,x])\,d\mu(x) = \int_0^1\int_0^x\,d\mu(y)\,d\mu(x) = \int_0^1\int_y^1\,d\mu(x)\,d\mu(y) = \int_0^1\mu([y,1])\,d\mu(y). $$ Therefore (note that $x$ and $y$ are dummy variables), \begin{align*} \int_0^1 \mu([0,x])\,d\mu(x) &= \frac{1}{2}\int_0^1\mu([0,x])+\mu([x,1])\,d\mu(x) \\ &= \frac{1}{2}\int_0^1\mu([0,1])\,d\mu(x) \\ &= \frac{1}{2}\mu([0,1])^2 \\ &= \frac{1}{2}m(A)^2. \end{align*} Note, we used that $\mu(\{x\}) = 0$ for every $x$ to say $\mu([0,x]) + \mu([x,1]) = \mu([0,1])$.