I have just finished the following problem for my measure theory course and wanted some feedback on my work:
Suppose that $f: [0,1] \to \mathbb{R}$ is defined by letting $f(x)=0$ on the Cantor set $\mathcal{C}$ and $f(x) =k$ for all $x$ in each interval of length $3^{-k}$ which has been removed from $[0,1]$. Calculate $\int^{1}_0fdm$ (the Lebesgue integral of $f$).
I recently posted about this question when I was working on it to gain further insight, but here is what I was able to come up with.
We have that $$f(x) = \sum_{k=1}^{\infty} k2^{k-1}\chi_{A_{k}}(x) \label{a}\tag{1}$$ where $\chi_{A_{k}}$ is the indicator function of $A_k$, the union of $2^{k-1}$ intervals of length $3^{-k}$ each, that are removed from $[0,1]$ at the $k^{th}$ stage. Since the convergence is monotone, we know that $$\int_{[0,1]}fdm=\lim_{n\to \infty} \sum_{k=1}^nk \frac{2^{k-1}}{3^{k}} = \frac{1}{3}\sum_{k=1}^{\infty}k(\frac{2}{3})^{k-1} \label{b}\tag{2}.$$ Since we know the following $$\sum_{k=1}^{\infty}k\alpha^{k-1}=\frac{1}{(1-\alpha)^2}$$ we can see that $\alpha = \frac{2}{3}$ in the equation $(2)$ so therefore $\int_{[0,1]}fdm = 3._{\Box}$
This all makes sense to me, however here is where I have questions:
$\mathrm{I})$ In my previous post (hyperlinked above), I was told that the $2^{k-1}$ term is wrong but my professor said it was correct. With that said, where does the $2^{k-1}$ actually come from? The reason I ask is my professor told me (via a hint) to define $f$ the way I did in $(1)$ (and i'm not sure I totally understand why $f$ is given this way).
$\mathrm{II})$ for the statement "since the convergence is monotone": I feel like I don't fully understand how I am using the monotone convergence theorem here and why the convergence is monotone. How can I explain this in more detail?
Any feedback is welcome.