Remove even elements of partition of integration set Suppose I am integrating a continuous $f:\mathbb{R}\rightarrow\mathbb{R}$
in a measurable set $A\subseteq I$, where $I$ is an interval:
$$
\int_{A}f(x)dx
$$
Now suppose I partition the set $I$ in $N$ intervals $I_{i}$ of
equal length: $I=\cup_{n=1}^{N}I_{n}$. Let $A_{n}=A\cap I_{n}$,
so that $\{A_{n}:n\in\{1,..,N\}\}$ is a partition of $A$. 
Let $E_N=\cup_{n=1}^{N/2}A_{2n}$
(it considers only the even elements of the partition of $A$). I
am trying to show that 
$$
\lim_{N\rightarrow\infty}\int_{E_N}f(x)dx=\frac{1}{2}\int_{A}f(x)dx
$$
But no success. I figured this may be a known result, is it?
 A: I think there may be a way to make this work. Define $B_N=\bigcup_{n=1}^{\lfloor N/2\rfloor} I_{2n}$ where $I_n$ is the $n$th interval (ordered by increasing left endpoint) partitioning $I$. Then define measures $\mu_n(A)=\int_A\chi_{B_n}\text d m$ on $I$. It is clear that $\mu_n$ are positive Radon measures with $\sup_n||\mu_n||=C<\infty$. By passing to a subsequence, we may assume that $\mu_n\rightharpoonup \mu\in M(I)$. 
It is straightforward (although tedious) to show that $\int_I x^n \text d\mu_n \to \frac 12\int_I x^n \text d m$, so the bounded linear functionals $\int_I f\text d\mu$ and $\frac 12\int f\text dm$ agree on the set of polynomials on $I$, hence they agree on $C(I)$.
Now, any measurable $A$ can be approximated by a compact set from within, so coupling a limiting argument with Urysohn's Lemma shows that $\int_A f\text d\mu=\frac 12\int_A f\text d m$.
I think this might be a way to formalize the idea in the question, although there may be other, more elegant or general paths.
A: I think the OP is trying to argue as follows.  Given $I=[0,1]$ one can construct a set $H\subset I$ by taking the limit of what you get when you divide $I$ into $n$ equal intervals and take the union of every other interval.  With this set in hand, $\lambda(A\cap H)=\frac 1 2 \lambda(A)$ for measurable $A$ (where $\lambda$ is Lebesgue measure) and $\int_H f(x)dx=\frac 1 2 \int_I f(x)dx$ for continuous $f$.
This program does not work.
As Guacho Perez points out, the limit $H$ is not precisely defined, and the OP has not supplied any argument that such a limit exists. 
 (It is certainly not a monotone limit.  It is somewhat like the supposed limit of the set of reals whose $n$-th decimal digit is even.)  And if it were such a set $H$, one would find that the measure $\nu\ll\lambda$ given by $\nu(A)=\lambda(A\cap H)$ has Radon-Nikodym derivative simultaneously equal to the constant function $1/2$ and to the indicator function of the set $H$. 
An alternative interpretation of what the OP is trying to say is:   Divide $I$ into $n$ equal subintervals, let $H_n$ be the union of every other one of them.  Then, for continuous bounded $f$, we have $\lim_{n\to\infty}\int_{H_n}f(x)dx = \frac 1 2 \int_I f(x)dx$.  This is true because the measure $\lambda_n$ given by $\lambda_n(A)=\lambda(A\cap I_n)/\lambda(I_n)$ converges weakly to $\lambda$.  Note here we have a limit of integrals, in contrast to the OP's original integral over a range defined by a limit.
Added next day, after edit of original problem.  The second interpretation above is what the OP intended, and the claimed result is correct. Namely, Lebesgue measure restricted to the OP's $E_N$, converges weakly to one half of Legbesgue
measure. There are many ways to check this.  (The open intervals are a "convergence determining class", and it is trivial to check that  $\lambda(E_N\cap(a,b))\to (b-a)/2$ for all $a<b$.)
