5
$\begingroup$

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 $\int\limits_{[0,1]}{{{\chi }_{\mathbb{Q}\cap [0,1]}}\left( x \right)}dx=0$.

$\endgroup$
3
  • $\begingroup$ Why exactly is $\Phi(x) = \lim_{n\to \infty} f_n(x) = 1$ only for rational numbers $x$? The set of numbers $x$ for which this limit is $1$ contains $U = \bigcap_{n=1}^\infty U_n$, where $U_n = (a_1 - \delta_{1,n}^-, a_1 + \delta_{1,n}^+) \cup \cdots \cup (a_n - \delta_{n,n}^-,a_n+\delta_{n,n}^+)$ is open. This set $U$ is a $G_\delta$-set containing $Q = \mathbb{Q} \cap [0,1]$ and is therefore strictly larger than $Q$. $\endgroup$
    – Martin
    Mar 16 '13 at 22:44
  • $\begingroup$ A further (minor) point is why the pointwise limit $\Phi(x) = \lim_{n \to \infty} f_n(x)$ exists in the first place. You can arrange this by making sure that $\delta_{i,n}^-$ and $\delta_{i,n}^+$ are monotonically decreasing functions of $n$. $\endgroup$
    – Martin
    Mar 16 '13 at 22:53
  • $\begingroup$ Please don't blank your questions. $\endgroup$
    – Potato
    Mar 27 '13 at 5:03
3
$\begingroup$

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 $\mathbb Q$.

$\endgroup$
1
  • $\begingroup$ Agreed. The way to make formal sense of these limits is to interpret them as $\lim_{n\to\infty}[a_i - \delta_{i,n}^-,a_i+\delta_{i,n}^+] = \bigcap_{n=i}^\infty[a_i-\delta_{i,n}^-,a_i+\delta_{i,n}^+]=\{a_i\}$ and then one sees that the argument involves the incorrect identification of $\bigcup_{i=1}^\infty \bigcap_{n=i}^\infty[a_i - \delta_{i,n}^-,a_i+\delta_{i,n}^+]$ with $\bigcap_{n=1}^\infty \bigcup_{i=1}^n [a_i - \delta_{i,n}^-,a_i + \delta_{i,n}^+]$. As I pointed out in my comments to the question, the latter set is strictly larger than the former (the former is $\mathbb{Q} \cap [0,1]$). $\endgroup$
    – Martin
    Mar 16 '13 at 23:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.