TITLE: Problem with the Lebesgue integral of Dirichlet function
QUESTION [6 upvotes]: According to all sources $\int\limits_{[0,1]}{{{\chi }_{\mathbb{Q}\cap [0,1]}}\left( x \right)}dx=0$ (Lebesgue integral out of Dirichlet funtion).
However, below I constructed a sequence of functions ${{f}_{n}}$ which converges to
$\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}(x)=\Phi (x)$ and ${{\chi }_{\mathbb{Q}\cap [0,1]}}\left( x \right)=\Phi (x)$
$\underset{n\to \infty }{\mathop{\lim }}\,\int\limits_{[0,1]}{{{f}_{n}}(x)}dx=\int\limits_{[0,1]}{\Phi (x)}dx=p\in (0,1)$
so $p$ is an arbitrary number from the interval (0,1). 
This is a contradiction because $\int\limits_{[0,1]}{{{\chi }_{\mathbb{Q}\cap [0,1]}}\left( x \right)}dx=0$.
Maybe somebody will be able to find a problem in my reasoning ...
The biggest problem in the method is the assumption that [\Phi (x)=0] for $x$ which is irrational number in the interval [0,1]. In my opinion I proved that it is true. Do you agree with my reasoning (see below for details).

Let us consider characteristic function of the set $\mathbb{Q}\cap [0,1]$ (Dirichlet funtion)
${{\chi }_{\mathbb{Q}\cap [0,1]}}\left( x \right)=\left\{ \begin{align}
  & 1\text{ for }x\in \mathbb{Q}\cap [0,1] \\ 
 & 0\text{ for }x\notin \mathbb{Q}\cap [0,1] \\ 
\end{align} \right.$
Let $\left\{ {{a}_{i}} \right\}_{i=1}^{\infty }$ be a sequence which contains all rational numbers i.e. $\mathbb{Q}=\bigcup\limits_{i=1}^{\infty }{\left\{ {{a}_{i}} \right\}}$.
Let’s define the following function
${{f}_{n}}(x)=\left\{ \begin{align}
  & 1\text{ if }\underset{i\in \{1,...,n\}}{\mathop{\exists }}\,x\in \left[ {{a}_{i}}-\delta _{i,n}^{-},{{a}_{i}}+\delta _{i,n}^{+} \right] \\ 
 & 0\text{ otherwise} \\ 
\end{align} \right.$
where 
$\delta _{i,n}^{+}-\delta _{i,n}^{-}={{\delta }_{n}}$ and $p=\sum\limits_{i=1}^{n}{{{\delta }_{n}}}=\sum\limits_{i=1}^{n}{\left( \delta _{i,n}^{+}-\delta _{i,n}^{-} \right)}$
and $p\in (0,1)$.
For every sequence of rational numbers $\left\{ {{a}_{i}} \right\}_{i=1}^{\infty }$ and $1>p>0$ it is possible to construct the intervals $\left[ {{a}_{i}}-\delta _{i,n}^{-},{{a}_{i}}+\delta _{i,n}^{+} \right]$ such that $p=\sum\limits_{i=1}^{n}{{{\delta }_{n}}}=\sum\limits_{i=1}^{n}{\left( \delta _{i,n}^{+}-\delta _{i,n}^{-} \right)}$ and $\left[ {{a}_{i}}-\delta _{i,n}^{-},{{a}_{i}}+\delta _{i,n}^{+} \right]\cap \left[ {{a}_{j}}-\delta _{j,n}^{-},{{a}_{j}}+\delta _{j,n}^{+} \right]=\varnothing $.
In order to do that first it is necessary to construct a sequence of intervals
$\left[ {{a}_{i}}-\varepsilon _{i,n}^{-},{{a}_{i}}+\varepsilon _{i,n}^{+} \right]$ such that $1=\sum\limits_{i=1}^{n}{{{\varepsilon }_{n}}}=\sum\limits_{i=1}^{n}{\left( \varepsilon _{i,n}^{+}-\varepsilon _{i,n}^{-} \right)}$ and $\left( {{a}_{i}}-\varepsilon _{i,n}^{-},{{a}_{i}}+\varepsilon _{i,n}^{+} \right)\cap \left( {{a}_{j}}-\varepsilon _{j,n}^{-},{{a}_{j}}+\varepsilon _{j,n}^{+} \right)=\varnothing $.
For any fixed $n>1$ it is possible to order all $\{{{a}_{1}},...,{{a}_{n}}\}$ such that ${{a}_{{{\alpha }_{1}}}}<{{a}_{{{\alpha }_{2}}}}<...<{{a}_{{{\alpha }_{n}}}}$.
Now we can define the following numbers
$\varepsilon _{1}^{-}={{a}_{{{\alpha }_{1}}}}$
$\varepsilon _{1}^{+}=\frac{{{a}_{{{\alpha }_{1}}}}+{{a}_{{{\alpha }_{2}}}}}{2}$
${{\varepsilon }_{1}}=\varepsilon _{1}^{-}+\varepsilon _{1}^{+}$
$\varepsilon _{2}^{-}=\frac{{{a}_{{{\alpha }_{1}}}}+{{a}_{{{\alpha }_{2}}}}}{2}$
$\varepsilon _{2}^{+}=\frac{{{a}_{{{\alpha }_{2}}}}+{{a}_{{{\alpha }_{3}}}}}{2}$
${{\varepsilon }_{2}}=\varepsilon _{2}^{-}+\varepsilon _{2}^{+}$
...
$\varepsilon _{n-1}^{-}=\frac{{{a}_{{{\alpha }_{n-2}}}}+{{a}_{{{\alpha }_{n-1}}}}}{2}$
$\varepsilon _{n-1}^{+}=\frac{{{a}_{{{\alpha }_{n-1}}}}+{{a}_{{{\alpha }_{n}}}}}{2}$
${{\varepsilon }_{n-1}}=\varepsilon _{n-1}^{-}+\varepsilon _{n-1}^{+}$
$\varepsilon _{n}^{-}=\frac{{{a}_{{{\alpha }_{n-1}}}}+{{a}_{{{\alpha }_{n}}}}}{2}$
$\varepsilon _{n}^{+}=1-{{a}_{{{\alpha }_{n}}}}$
${{\varepsilon }_{n}}=\varepsilon _{n}^{-}+\varepsilon _{n}^{+}$
It is possible to see that
$\sum\limits_{i=1}^{n}{{{\varepsilon }_{i}}}=1$
$\sum\limits_{i=1}^{n}{{{\varepsilon }_{i}}}={{\varepsilon }_{1}}+{{\varepsilon }_{2}}+...+{{\varepsilon }_{n-1}}+{{\varepsilon }_{n}}=(\varepsilon _{1}^{-}+\varepsilon _{1}^{+})+(\varepsilon _{2}^{-}+\varepsilon _{2}^{+})+...+(\varepsilon _{n-1}^{-}+\varepsilon _{n-1}^{+})+(\varepsilon _{n}^{-}+\varepsilon _{n}^{+})=$
$=\left( {{a}_{{{\alpha }_{1}}}}+\frac{{{a}_{{{\alpha }_{2}}}}-{{a}_{{{\alpha }_{1}}}}}{2} \right)+\left( \frac{{{a}_{{{\alpha }_{2}}}}-{{a}_{{{\alpha }_{1}}}}}{2}+\frac{{{a}_{{{\alpha }_{3}}}}-{{a}_{{{\alpha }_{2}}}}}{2} \right)+\left( \frac{{{a}_{{{\alpha }_{3}}}}-{{a}_{{{\alpha }_{2}}}}}{2} \right.+...+\left. \frac{{{a}_{{{\alpha }_{n-1}}}}-{{a}_{{{\alpha }_{n-2}}}}}{2} \right)+$
$+\left( \frac{{{a}_{{{\alpha }_{n-1}}}}-{{a}_{{{\alpha }_{n-2}}}}}{2}+\frac{{{a}_{{{\alpha }_{n}}}}-{{a}_{{{\alpha }_{n-1}}}}}{2} \right)+\left( \frac{{{a}_{{{\alpha }_{n}}}}-{{a}_{{{\alpha }_{n-1}}}}}{2}+1-{{a}_{{{\alpha }_{n}}}} \right)=$
$={{a}_{{{\alpha }_{1}}}}+\frac{{{a}_{{{\alpha }_{2}}}}-{{a}_{{{\alpha }_{1}}}}}{2}+\frac{{{a}_{{{\alpha }_{2}}}}-{{a}_{{{\alpha }_{1}}}}}{2}+\frac{{{a}_{{{\alpha }_{3}}}}-{{a}_{{{\alpha }_{2}}}}}{2}+\frac{{{a}_{{{\alpha }_{3}}}}-{{a}_{{{\alpha }_{2}}}}}{2}+...$
$+\frac{{{a}_{{{\alpha }_{n-1}}}}-{{a}_{{{\alpha }_{n-2}}}}}{2}+\frac{{{a}_{{{\alpha }_{n-1}}}}-{{a}_{{{\alpha }_{n-2}}}}}{2}$
$+\frac{{{a}_{{{\alpha }_{n}}}}-{{a}_{{{\alpha }_{n-1}}}}}{2}+\frac{{{a}_{{{\alpha }_{n}}}}-{{a}_{{{\alpha }_{n-1}}}}}{2}+1-{{a}_{{{\alpha }_{n}}}}=$
$={{a}_{{{\alpha }_{1}}}}-\frac{{{a}_{{{\alpha }_{1}}}}}{2}+\frac{{{a}_{{{\alpha }_{2}}}}}{2}-\frac{{{a}_{{{\alpha }_{2}}}}}{2}+\frac{{{a}_{{{\alpha }_{1}}}}}{2}-\frac{{{a}_{{{\alpha }_{3}}}}}{2}+\frac{{{a}_{{{\alpha }_{3}}}}}{2}+...$
$+\frac{{{a}_{{{\alpha }_{n-2}}}}}{2}-\frac{{{a}_{{{\alpha }_{n-1}}}}}{2}+\frac{{{a}_{{{\alpha }_{n-1}}}}}{2}-\frac{{{a}_{{{\alpha }_{n-2}}}}}{2}+\frac{{{a}_{{{\alpha }_{n}}}}}{2}-\frac{{{a}_{{{\alpha }_{n-1}}}}}{2}+\frac{{{a}_{{{\alpha }_{n}}}}}{2}-\frac{{{a}_{{{\alpha }_{n-1}}}}}{2}+1-{{a}_{{{\alpha }_{n}}}}=$
$={{a}_{{{\alpha }_{1}}}}-{{a}_{{{\alpha }_{1}}}}+{{a}_{{{\alpha }_{2}}}}-{{a}_{{{\alpha }_{2}}}}+...+{{a}_{{{\alpha }_{n}}-1}}-{{a}_{{{\alpha }_{n-1}}}}+{{a}_{{{\alpha }_{n}}}}+1-{{a}_{{{\alpha }_{n}}}}=1$
In every interval $\left[ {{a}_{i}}-\varepsilon _{i,n}^{-},{{a}_{i}}+\varepsilon _{i,n}^{+} \right]$ it is possible to find subintervals $\left[ {{a}_{i}}-\delta _{i,n}^{-},{{a}_{i}}+\delta _{i,n}^{+} \right]$ such that $\left[ {{a}_{i}}-\delta _{i,n}^{-},{{a}_{i}}+\delta _{i,n}^{+} \right]\subset \left[ {{a}_{i}}-\varepsilon _{i,n}^{-},{{a}_{i}}+\varepsilon _{i,n}^{+} \right]$ and
$p=\sum\limits_{i=1}^{n}{l\left( \left[ {{a}_{i}}-\delta _{i,n}^{-},{{a}_{i}}+\delta _{i,n}^{+} \right] \right)}=\sum\limits_{i=1}^{n}{\left( \delta _{i,n}^{+}-\delta _{i,n}^{-} \right)}=\sum\limits_{i=1}^{n}{{{\delta }_{i,n}}}$.
In order to do that it is enough to assume that ${{\delta }_{i,n}}=p{{\varepsilon }_{i,n}}$ or $\delta _{i,n}^{-}=p\varepsilon _{i,n}^{-},\delta _{i,n}^{+}=p\varepsilon _{i,n}^{+}$.
Verification
$\sum\limits_{i=1}^{n}{{{\varepsilon }_{i}}}=1$
$\sum\limits_{i=1}^{n}{{{\delta }_{i}}}=\sum\limits_{i=1}^{n}{p{{\varepsilon }_{i}}}=p\sum\limits_{i=1}^{n}{{{\varepsilon }_{i}}}=p1=p$
additionally because $p\varepsilon _{i,n}^{-}<\varepsilon _{i,n}^{-}$ and $p\varepsilon _{i,n}^{+}<\varepsilon _{i,n}^{+}$ then
$\left[ {{a}_{i}}-\delta _{i,n}^{-},{{a}_{i}}+\delta _{i,n}^{+} \right]=\left[ {{a}_{i}}-p\varepsilon _{i,n}^{-},{{a}_{i}}+p\varepsilon _{i,n}^{+} \right]\subset \left[ {{a}_{i}}-\varepsilon _{i,n}^{-},{{a}_{i}}+\varepsilon _{i,n}^{+} \right]$.
Because of that for each $n$
$\int\limits_{[0,1]}{{{f}_{n}}(x)dx}=\sum\limits_{i=1}^{n}{{{\delta }_{n}}}=p$
then 
$\underset{n\to \infty }{\mathop{\lim }}\,\int\limits_{[0,1]}{{{f}_{n}}(x)dx}=\underset{n\to \infty }{\mathop{\lim }}\,p=p\in (0,1)$
Because set $\mathbb{Q}\cap [0,1]$ is dense in [0,1] then $\underset{n\to \infty }{\mathop{\lim }}\,\delta _{i,n}^{-}=0$, $\underset{n\to \infty }{\mathop{\lim }}\,\delta _{i,n}^{+}=0$.
For every fixed ${{a}_{i}}$ we have
$\underset{n\to \infty }{\mathop{\lim }}\,\left[ {{a}_{i}}-\delta _{i,n}^{-},{{a}_{i}}+\delta _{i,n}^{+} \right]=\left[ {{a}_{i}}-\underset{n\to \infty }{\mathop{\lim }}\,\delta _{i,n}^{-},{{a}_{i}}+\underset{n\to \infty }{\mathop{\lim }}\,\delta _{i,n}^{-} \right]=\left[ {{a}_{i}}-0,{{a}_{i}}+0 \right]=\{{{a}_{i}}\}$
$\underset{n\to \infty }{\mathop{\lim }}\,\bigcup\limits_{i=1}^{n}{\left[ {{a}_{i}}-{{\delta }_{n}},{{a}_{i}}+{{\delta }_{n}} \right]}=\{{{a}_{i}}\}_{i=1}^{\infty }=\mathbb{Q}\cap [0,1]$
Let
$\Phi (x)=\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}(x)$
and x is a rational nuber in the interval [0,1] then
$\Phi (x)=\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}(x)=1$   for   $x\in \mathbb{Q}\cap [0,1]$
In limit case $\underset{n\to \infty }{\mathop{\lim }}\,\delta _{i,n}^{-}=\underset{n\to \infty }{\mathop{\lim }}\,\delta _{i,n}^{-}=0$ then function $\Phi (x)=\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}(x)$ is 1 only for rational numbers consequently $\Phi (x)=\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}(x)=0$ for x which is an irrational number in the interval [0,1] i.e.
$x\in R\backslash \left( \mathbb{Q}\cap [0,1] \right)$ then $\Phi (x)=\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}(x)=0$
Because of that 
$\Phi (x)=\underset{n\to \infty }{\mathop{\lim }}\,{{f}_{n}}(x)={{\chi }_{\mathbb{Q}\cap [0,1]}}\left( x \right)$
According to the definition of the Lebesgue integral we have
$\underset{n\to \infty }{\mathop{\lim }}\,\int\limits_{[0,1]}{{{f}_{n}}(x)}dx=\int\limits_{[0,1]}{{{\chi }_{\mathbb{Q}\cap [0,1]}}\left( x \right)}dx=p\in (0,1)$
which is a contradiction because $\int\limits_{[0,1]}{{{\chi }_{\mathbb{Q}\cap [0,1]}}\left( x \right)}dx=0$.

REPLY [3 votes]: Your mistake (in an otherwise nice attempt) seems to be in the following lines:
"For every fixed ${{a}_{i}}$ we have
$\lim_{n\to \infty} \left[ {{a}_{i}}-\delta _{i,n}^{-},{{a}_{i}}+\delta _{i,n}^{+} \right]=\left[ {{a}_{i}}-\lim_{n\to \infty} \delta _{i,n}^{-},{{a}_{i}}+\lim_{n\to \infty} \delta _{i,n}^{-} \right]=\left[ {{a}_{i}}-0,{{a}_{i}}+0 \right]=\{{{a}_{i}}\}$
$\lim_{n\to \infty} \bigcup\limits_{i=1}^{n}{\left[ {{a}_{i}}-{{\delta }_{n}},{{a}_{i}}+{{\delta }_{n}} \right]}=\{{{a}_{i}}\}_{i=1}^{\infty }=\mathbb{Q}\cap [0,1]$"
You are right that you want to calculate $\lim_{n\to \infty} \bigcup\limits_{i=1}^{n}{\left[ {{a}_{i}}-{{\delta }_{n}},{{a}_{i}}+{{\delta }_{n}} \right]}$. However, what you have actually calculated (in the rest of that line) is $\lim_{n\to \infty} \bigcup\limits_{i=1}^{n} \big( \lim_{m\to \infty} {\left[ {{a}_{i}}-{{\delta }_{m}},{{a}_{i}}+{{\delta }_{m}} \right]} \big)$, which is not the same thing.
Here's one way to see that your construction isn't what you think it is: your argument was for some $p\in(0,1)$, but note that it is completely the same for $p=1$ as well. In this case, each $f_n$ is simply $1$ on all of $[0,1]$, and hence so is $\lim_{n\to\infty} f_n$. But in the mistake I copied above, the logic would conclude equally well that $\lim_{n\to\infty} f_n$ is the indicator function of $\mathbb Q$.