Question

What is the value of this integral? \[\int\limits_{\mathbb Q \cap [0,1]} dx\]Here, \(\mathbb Q \cap [0,1]\) is the set of rational numbers between 0 and 1, inclusive (I'm essentially finding the total "length" of the set of rational numbers on a number line). My initial thoughts were that this integral would be zero, since there are no discrete intervals, though, if you think about it,\[\int\limits_{\mathbb Q \cap [0,1]} dx + \int\limits_{[0,1]-\mathbb Q} dx \equiv \int\limits_{[0,1]} dx = 1\]Integrating over \([0,1]-\mathbb Q\) (irrationals) also is weird, but one must be nonzero!

Answer 1

Any thoughts or input would be appreciated!

Answer 2

dont you need to define an integral over an interval? not just a collection of points?

Answer 3

No...think of this as the integral over a region, like you do all the time in multivariable calculus, or, if you prefer:\[\int\limits_{\mathbb Q \cap [0,1]} dx\equiv \int\limits_0^1 f(x)\ dx\]where\[f(x) = \left\{ \begin{array}{ll} 1 & \text{if } x \in \mathbb Q \\ 0 & \text{if } x \not \in \mathbb Q\end{array}\right.\]

Answer 4

thats what i mean, you changed it to an integral over the interval [0,1], and made the function have the value 1 over rationals, and 0 over irrationals. Your initial problem didnt state what the value of the function was over irrationals. It was as if you were ignoring them completely.

Answer 5

The value of the original integrand was 1 (\(1\ dx = dx\)), over the set. They're equivalent.

Answer 6

That's how the definition of \(f(x)\) comes about...it's only 1 in the region, and doesn't contribute anything when out of the region.

Answer 7

Although my Real Analysis isnt very sharp, i believe that function is not Riemann Integrable.

Answer 8

I found a section of my analysis book that talks about it. I would prbably butcher it if I tried to type it out. Gotta study Analysis more.

Answer 9

Interesting...thanks for that.... and it's also interesting to know that there's a name to go along with my function lol.