1. ## Integrability

Let $\displaystyle A\subseteq[0,1], |A|=\aleph_0$,

f(x)={0 if $\displaystyle x\notin A$, 1 if $\displaystyle x\in A$}

Prove that f is not integrable in [0,1], or show a negative proof and prove that it is integrable in [0,1].

Let $\displaystyle A\subseteq[0,1], |A|=\aleph_0$,

f(x)={0 if $\displaystyle x\notin A$, 1 if $\displaystyle x\in A$}

Prove that f is not integrable in [0,1], or show a negative proof and prove that it is integrable in [0,1].
Reimann integrable or Lebesque integrable? In either case, since f is bounded and its set of discontinuities has Lebesque measure 0, f is integrable.

3. Originally Posted by HallsofIvy
Reimann integrable or Lebesque integrable? In either case, since f is bounded and its set of discontinuities has Lebesque measure 0, f is integrable.
Reimann integrable. I haven't learned Lebesque measure - only Reimann and Darboo's integrable definition.

4. Originally Posted by HallsofIvy
Reimann integrable or Lebesque integrable? In either case, since f is bounded and its set of discontinuities has Lebesque measure 0, f is integrable.
and BTW in Riemann's case f is not integrable - take Dirichlet function in [0,1] - it is not Riemann-integrable, because you can always take a group of irrational points to get an integral that equals to zero, and so on.

Hint:

Spoiler:
$\displaystyle N=\left\{\frac{1}{n}:n\in\mathbb{N}\right\}\cup\{0 \}$

Spoiler:
Let $\displaystyle \varepsilon>0$ be given. Clearly $\displaystyle [0,1]-N$ is dense and so $\displaystyle L(P,f)=0$ for every parition $\displaystyle P$. So, we must merely show that we may make $\displaystyle U(P,f)$ arbitrarily small. So, the idea is to cover all but finitely many elements of $\displaystyle N$ by one interval containing $\displaystyle 0$ and then put sufficiently small intervals around the remaining points. I have distictly left out three important details. Fill them in.

7. Originally Posted by HallsofIvy
Reimann integrable or Lebesque integrable? In either case, since f is bounded and its set of discontinuities has Lebesque measure 0, f is integrable.
Just because your perturbing the zero function at a countable number of places does not mean the set of all discontinuities created is countable, and thus necessarily of measure zero. The indicator function for the rationals differs from the zero function at countably many places but the set of discontinuities introduced has measure one.