Lebesgue integral and partition of an interval For any Lebesuge integrable function f on $[0,1]$, it holds
$$ \lim_n \sum_{k=0}^{n-1} \int_{\frac{2k}{2n}}^{\frac{2k+1}{2n}} f(x) \,dx = \frac{1}{2}\int_0^1 f(x) \, dx.$$
For example, when $n =3$, the left term is
$$\int_{o}^1 \left(1_{[\frac{0}{6}, \frac{1}{6}]}+1_{[\frac{2}{6}, \frac{3}{6}]}+1_{[\frac{4}{6}, \frac{5}{6}]} \right)(x) f(x) \, dx.$$
Intuitively, it is obvious because the interval of integration will be devided into small pieces, total measure of which is $\frac{1}{2}.$
I`m thinking of using dominated convergence theorem by setting $f_n$ , but it does not converge to $\frac{1}{2}f$. Please help me.
 A: I don't think that using DCT will work here - as you said, the convergence does not hold. One option is to do it by hand: for indicator functions, then simple ones, and then in general. First, assume that $f$ is an indication function on some interval $(a, b)$. Then it is not so difficult to convince ourselves that the result holds: divide $(0, 1)$ in $2n$ intervals, and see how the partition restricts to $(a, b)$: the difference between the invervals $(2k/2n, 2k+1/2n)$ and $(2k-1/2n, 2k/2n)$ is at most $1/n$ I think.
The next step is to show it for arbitrary indicator functions, for which is convenient to establish in this more general setting: if $g_s(x)$ are functions that already satisfies the result, and $\int_0^1 |f-g_s| \to 0$, then the result holds for $f$ as well. (This is enough: if $f$ is the indicator of some set $E$, use that there are some open set $U_s \supseteq E$ such that $\mu(U_s\setminus E) \to 0$.) In fact, we do the trick
\begin{align*}
\left| \sum_{k=0}^{n-1} \int_{\frac{2k}{2n}}^{\frac{2k+1}{2n}} f(x)\,\mathrm{d}x- \frac{1}{2} \int_0^1 f(x) \,\mathrm{d}x \right| \leq& \left| \sum_{k=0}^{n-1} \int_{\frac{2k}{2n}}^{\frac{2k+1}{2n}} g_s(x)\,\mathrm{d}x- \frac{1}{2} \int_0^1 g_s(x) \,\mathrm{d}x \right| \\
&+ \sum_{k=0}^{n-1}  \int_{\frac{2k}{2n}}^{\frac{2k+1}{2n}} |f(x)-g_s(x)| \,\mathrm{d}x + \frac{1}{2} \int_0^1 |f(x)-g_s(x)| \,\mathrm{d}x \\
\leq& \left| \sum_{k=0}^{n-1} \int_{\frac{2k}{2n}}^{\frac{2k+1}{2n}} g_s(x)\,\mathrm{d}x- \frac{1}{2} \int_0^1 g_s(x) \,\mathrm{d}x \right| \\
&+ \frac{3}{2} \int_0^1 |f(x)-g_s(x)| \,\mathrm{d}x.
\end{align*}
(If you are more careful I think it is possible to replace the 3/2 with a 1/2, but this is not relevant.)
This way, now fix some $\varepsilon>0$. There is some $s$ such that $\int_0^1 |f(x)-g_s(x)| \,\mathrm{d}x<\frac{1}{3}\varepsilon$. Then, as we already proved the result for $g_s$, there is some $n_0$ such that
$$ \left| \sum_{k=0}^{n-1} \int_{\frac{2k}{2n}}^{\frac{2k+1}{2n}} g_s(x)\,\mathrm{d}x- \frac{1}{2} \int_0^1 g_s(x) \,\mathrm{d}x \right| < \frac{\varepsilon}{2}, \qquad n \geq n_0. $$
This implies that
$$ \left| \sum_{k=0}^{n-1} \int_{\frac{2k}{2n}}^{\frac{2k+1}{2n}} f(x)\,\mathrm{d}x- \frac{1}{2} \int_0^1 f(x) \,\mathrm{d}x \right| < \varepsilon, \qquad n \geq n_0, $$
and so we get the convergence.
This looks a bit scary, but the idea is not so difficult: we show the result for some ``nice'' functions $g$, and then we approximate to prove it in general. (If you have seen the proof of the Riemann-Lebesgue lemma for instance, the trick is similar.)
The rest of the proof follows the same ideas: we prove it for simple functions by linearity, and then we use that every integrable functions can be approximated by simple functions. (And so we apply the proof with $f$ and $g_s$ from above.)
