Only Transcendentals Known

$\mathbb {R} - \mathbb{T}$ are the algebraic numbers, which are roots of polynomials with integer coefficients. We can show that there are only a Countable number of these, hence the algebraic numbers have a measure of 0.

Since we're working with lebesgue integral , the value of the integral is independent of the value on a set of measure 0. As such, the integral can be calculated just over $\mathbb{T}$ , and we obtain that the value is 0.

Hence, for all functions $f$ that satisfy $f(t) = 0$ for transcedental $t$ (and have no restriction otherwise), we have $\int f = 0$ .

Note: All functions that satisfy $f(t) = 0$ for transcedental $t$ are Lebesgue integrable.

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If the question used the improper riemann integral instead, then the proof is slightly more complicated. We have to work from first principles, using the fact that the integral exists.

First, for the improper integral $\int_{-\infty} ^ \infty$ which is interpreted as $\lim_{(a,b) \rightarrow ( - \infty, \infty) } \int_a ^ b$ , we get that $f : [a,b] \rightarrow \mathbb{R}$ is Riemann-integrable.

Second, we will show that for fixed $a,b$ , we have $\int_a^b f(x)\, dx = 0$ . This follows from the definition of the Riemann integral:

For all $\epsilon > 0$ , there exists a $\delta$ such that for any tagged partition $x_0, \ldots , x_n$ and $t_0, \ldots , t_{n-1}$ whose mesh is less than $\delta$ , we have
$| \sum_{i=0}^{n-1} f(t_i) ( x_{i+1} - x_i) - S | < \epsilon$

For any $\epsilon > 0$ , there exists a $\delta_ \epsilon$ such that for all tagged partitions with mesh less than $\delta_\epsilon$ , $| \sum_{i=0}^{n-1} f(t_i) ( x_{i+1} - x_i) - S | < \epsilon$ . Fixing the partition to use internal interval endpoints of the form $n \delta_\epsilon$ and picking transedentals $t_i$ as our tags, we see that $\sum_{i=0}^{n-1} f(t_i) ( x_{i+1} - x_i) = 0$ . This gives us $|S | < \epsilon$ . Now, taking limits as $\epsilon \rightarrow 0$ , we conclude that $S = 0$ .

Finally, $\int_{-\infty} ^ \infty = \lim_{(a,b) \rightarrow ( - \infty, \infty) } \int_a ^ b = \lim 0 = 0$ .

Note: In contrast to the Lebesgue integral, there are a lot of functions which satisfy $f(t) = 0$ for transcendentals $t$ , but are not Riemann integrable. This includes functions which are unbounded on a compact interval.

Note: An alternative is to use the Riemann-Lebesgue criterion in the first half to conclude that we can perform the Lebesgue integral which yields a value of 0.