I am new to measure theory and I'm have a problem finding an example for the problem below:
If $C$ is a middle-thirds Cantor set; find a sequence $\{ \mu_{n}\}_{n \geq 1 }$ of measures on Borel $\sigma -$ algebra $C$ and a measurable function $f:C \to R$ such that $\lim_{n \to +\infty} \mu_n(x) = 0$ and $\int f d\mu_{m} = +\infty$ for every $m \geq 1$.
I think that this is the Cantor measure problem but I'm not sure.
Best Answer
I'm going to use the space of infinite binary strings (with the product topology) rather than the middle thirds cantor space. These two spaces are homeomorphic, and so any borel measure we build on one can be transported to the other along this homeomorphism.
Let $\mu_0$ be the standard coin-tossing measure on cantor space. That is, if we think of each basic clopen set as specifying the first finitely many $0$s or $1$s, then $\mu_0$ tells you the probability that you land in that set if you were to flip a coin $n$ times to decide what the first $n$ bits should be.
Let $\mu_n$ be the coin tossing measure conditioned on the event "the first $n$ bits are $0$".
Let's take a moment to see some examples. We'll write $\mathcal{N}_{s}$ for the set of strings starting with $s$.
$\mu_0 \mathcal{N}_{0010} = \frac{1}{16}$, since if we flip $4$ coins to determine our first $4$ bits, we get $0010$ with probability $\frac{1}{16}$.
$\mu_1 \mathcal{N}_{0010} = \frac{1}{8}$, since we're assuming the first bit is a $0$, but we still need to flip $3$ coins correctly to end up in this set.
$\mu_2 \mathcal{N}_{0010} = \frac{1}{4}$, since now we're assuming the first $2$ bits are $0$s, and we need to flip $2$ coins correctly to end up in this set.
$\mu_3 \mathcal{N}_{0010} = 0$, since we're assuming the first $3$ bits are $0$, which isn't the case for our set. Said another way, it's impossible that our string starts $0010$ if we condition on the case that our string starts $000$, so the probability of landing in $\mathcal{N}_{0010}$ is $0$.
It should be clear that $\lim \mu_n = 0$, since every point besides the all $0$s string is eventually excluded from the support of our measures.
Now let's define $f$. Let's put $f(s) = 2^k$, where $k$ is the index of the first $1$ in $s$.
Again, by examples,
Do you see why this function is measurable?
Notice $f$ is technically undefined at the all $0$s string, but that's a set of measure $0$, so ¯\_(ツ)_/¯. If you like, we can define $f(00000...) = 0$ to avoid this problem. The function will still be measurable and will do what we need it to do.
Ok, we're in the home stretch. Why does $\int f \ d\mu_n = \infty$ for each $n$?
Well, notice $f(s) = 2^{k+1}$ for every $s \in \mathcal{N}_{0^k 1}$. That is, $f$ is a countable sum of rescaled characteristic functions,
$$ f = 2 \chi_{\mathcal{N}_{1}} + 4 \chi_{\mathcal{N}_{01}} + 8 \chi_{\mathcal{N}_{001}} + \ldots = \sum 2^{k+1} \chi_{\mathcal{N}_{0^k 1}} $$
But we know that
$$ \mu_n \mathcal{N}_{0^k 1} = \begin{cases} \frac{1}{2^{k+1-n}} & k \geq n \\ 0 & k < n \end{cases} $$
and so (by the monotone convergence theorem) we find
$$ \begin{aligned} \int f \ d\mu_n &= \sum_k \int 2^{k+1} \chi_{\mathcal{N}_{0^k 1}} \ d\mu_n \\ &= \sum_k 2^{k+1} \mu_n \mathcal{N}_{0^k 1} \\ &= \sum_{k \geq n} 2^{k+1} \frac{1}{2^{k+1-n}} \\ &= \sum_{k \geq n} 2^n \\ &= \infty \end{aligned} $$
I hope this helps ^_^