Show that the function f, defined on 𝐈 by f(x) =x, x rational, =0,

step 1 We know that the interval i over n comma i plus 1 over n will always have a rational and irratio

Show that the function f, defined on 𝐈 by f(x) =x, ...

Show that the function $f$, defined on $\mathbf{I}$ by
$$
\begin{aligned}
f(x) & =x, & & x \text { rational, } \\
& =0, & & x \text { irrational, }
\end{aligned}
$$
is not Riemann integrable on $\mathbf{I}$.

Key Concepts

Discontinuities and Measure Zero Criterion

A crucial criterion for Riemann integrability is that the set of points where a function is discontinuous must have Lebesgue measure zero. If a function is discontinuous on an interval of nonzero measure – for instance, at every point – then the difference between the upper and lower sums cannot be made arbitrarily small, and the function is not Riemann integrable.

Density of Rational and Irrational Numbers

In the context of real numbers, both rational and irrational numbers are dense. This means that within any subinterval, no matter how small, both types of numbers can be found. When a function is defined differently on these two dense sets, it can lead to significant variations in its behavior on every subinterval, affecting the consistency required for integration.

Riemann Integrability

This concept refers to the idea of defining the integral of a function via limits of Riemann sums. A function is Riemann integrable on an interval if, for every ? > 0, there exists a partition such that the difference between the upper and lower sums is less than ?. The fundamental requirement is that the two sums converge to the same limit as the partitions become arbitrarily fine.

Upper and Lower Sums

These sums are used in the definition of the Riemann integral. The upper sum is calculated by taking, on each subinterval of a partition, the supremum of the function and multiplying it by the length of the subinterval, whereas the lower sum uses the infimum. A function is Riemann integrable only if the upper and lower sums converge to the same value as the partition is refined.

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