On the integration of a Lebesgue measurable function Consider a function $f$ defined as $f:[0,2\pi]\to \mathbb{R}$ such that 
$\begin{equation}
f(x)=\inf_{n\in \mathcal{N}} \sin^2 (2^n x)
\end {equation}$
Is possible to give a decent bound of $\int_{0}^{2\pi} f(x) dx$ or directly get its value?
We can prove that this function is measurable by prove the following lemma:
Lemma: If $\mathcal{F}$ is a family of continuous functions on $[0, 2\pi]$. Then functon $\phi(x)$ defined as $\phi(x)=\inf_{f\in \mathcal{F}} f(x)$
is measurable on $[0, 2\pi]$.
Hence, according to Lusin's theorem we can prove that this function is continuous a.e. on $[0,2\pi]$. 
Given all of these, can we get the exact value of the integration if it exists?
Some of my own ideas:
1) I think one of the hard point is to determine the structure of sets $\mathbb{E}_{t}=f^{-1}(t)$ for any given $t$ in the range of $f$.
2) another interpretation on $f$: notice that if we denote $b_{k}= \sin^2 (2^{k} x)$ for fixed $x$ then it's not too hard to check that  $b_{k+1}=4b_{k}(1-b_{k})$. Hence sequence $\{b_{k}(x)\}$ could be regarded as obtained by starting at any point $b_{0}(x)=sin^2(x), x\in [0,2\pi]$ and conduct iteration with respect of $g(x)=4x(1-x)$. $f(x)$ actually measures the infimum of sequence $\{b_{k}(x)\}$ regarding different values of $b_{0}(x), x\in [0,2\pi]$.
 A: I claim that $f=0$ almost everywhere.  To prove this, we first need a lemma.

Lemma: Let $A\subset [0,1]$ be the set of numbers $x$ such that every finite string of $0$s and $1$s appears somewhere in the binary expansion of $x$.  Then $A$ has measure $1$.
Proof:  There are only countably many such finite strings, so it suffices to show that for any finite string $s$, the set $A_s\subset[0,1]$ of numbers whose binary expansion contains $s$ as a substring has measure $1$.  Intuitively, this should make sense from a probabilistic perspective: if you think of the binary expansion as being obtained by an infinite sequence of coinflips, you should expect to eventually see any finite sequence with probability $1$.
Let's prove this rigorously.  Let $n$ be the length of $s$.  The set $B_1$ of numbers $x=0.x_1x_2x_3\dots$ such that $x_1x_2\dots x_n\neq s$ (i.e., the first $n$ digits of $x$ are not $s$) has measure $(2^n-1)/2^n$.  The set $B_2$ of numbers $x$ such that $x_1x_2\dots x_n\neq s$ and also $x_{n+1}x_{n+2}\dots x_{2n}\neq s$ has measure $(2^n-1)^2/2^{2n}$.  In general, if $B_k$ is the set of numbers $x$ such that $s$ is not equal to any of the first $k$ disjoint blocks of $n$ digits of $x$, $B_k$ will have measure $(2^n-1)^k/2^{kn}$.  Since $(2^n-1)^k/2^{kn}\to 0$ as $k\to\infty$, the intersection $\bigcap B_k$ must have measure $0$.  But this intersection contains the complement $[0,1]\setminus A_s$, so $A_s$ must have measure $1$.

Now let's use the lemma to show $f=0$ almost everywhere.  Given $x\in [0,2\pi]$, let $y=x/2\pi$.  Note that for any $n$, the binary expansion of the fractional part of $2^n y$ is just the binary expansion of $y$ with the first $n$ digits removed.  In particular, if $y\in A$, then the set of fractional parts of numbers of the form $2^n y$ is dense in $[0,1]$ (since any finite string is the start of the fractional part of $2^ny$ for some $n$).  This means that the values of $2^n x$ modulo $2\pi$ are dense in $[0,2\pi]$.  It follows that the values of $\sin^2 (2^n x)$ are dense in $[0,1]$.  In particular, the infimum of such values is $0$.  Thus $f(x)=0$ whenever $y\in A$, and it follows from the lemma that $f(x)=0$ almost everywhere.
