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Let $$f(x,y)= \begin{cases} 1,&\mbox{ if $x$ is rational}\\2y,&\mbox{ if $x$ is irrational.} \end{cases} $$

Show that $$\int_0^1dx\int_0^1f(x,y)dy=1$$

But that

$$\int_0^1dy\int_0^1f(x,y)dx$$

Fails to exist.

My approach : I couldn't manage to build a good approach here to be honest, but I think f(x,y)=1 if x is rational is not continuous that's why the second integral does not exist. But I also couldn't evaluate the first integral, any hints?

Juniven Acapulco
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    Is this Riemann integration? –  Dec 27 '16 at 16:22
  • Yes rıemann ıntegra –  Dec 27 '16 at 16:41
  • My guess is that, since the irrationals are dense in the reals, rationals can be thought of as removable discontinuities, which gives the first result. – Kaynex Dec 27 '16 at 16:53
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    The second (inner) integral doesn't exist by the same reason that the Riemann integral of any similar comb function doesn't exist, i.e. $g(x)=1$ if $x$ is rational and 0 otherwise. Just fix $y$ and prove the upper/lower sums do not converge to each other. – Alex R. Dec 27 '16 at 16:57
  • The first integral exists because, for a fixed $x$, the function is either $1$ or $2y$. The integral of both of these on the interval $[0,1]$ is $1$. – Steve D Dec 27 '16 at 17:46

2 Answers2

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Let's consider the first integral,

$$\int_0^1 \left( \int_0^1 f(x,y) dy\right) dx. \tag{1}$$

The inner integral is $$ \int_0^1 f(x,y) dy.$$ If $x$ is a rational number, then this is $$ \int_0^1 1 dy = 1.$$ If $x$ is irrational, then this integral is $$ \int_0^1 2y dy = y^2 \bigg|_0^1 = 1.$$ So the inner integral is always $1$. Thus we can rewrite $(1)$ as $$ \int_0^1 1 dx = 1.$$

Let us now consider the second integral, $$ \int_0^1 \left( \int_0^1 f(x,y) dx \right) dy. \tag{2}$$ We must consider the inner integral again. The inner integral is $$ \int_0^1 f(x,y) dx.$$ If $y$ is any number other than $\frac{1}{2}$, then this integral is not integrable. In short, as the rationals and irrationals are each dense in $[0,1]$, any upper sum partition of $[0,1]$ will yield a very different area estimate compared to any lower sum partition of $[0,1]$.

This function is essentially the Dirichlet Function, and it is a classical exercise to show that the Dirichlet Function is not integrable [See this answer on this site to a similar question for one explanation as to why $(2)$ is not integrable].

Since the inner integral in $(2)$ is not integrable, we have that this integral fails to exist.

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davidlowryduda
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The first integral exists because if we fix $x$ then the inner integrand can take either a value of $1$ or $2y$ and for both, the overall integral exists and comes out to be $1$. However for the second integral,it's inner integral can be thought of as an integration for rational indicator function,except in this case value is not $0$ when $x$ is irrational but $2y$, which doesn't exist (proved using density arguments).

P.S. Riemann integration is considered.

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Sai Kashyap
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