How to find the upper Riemann integral of following function?

Question

$f(x)$ is defined on $[a,b]$ as -

$= 0$ if $x ∈ [a, b] ∩ Q$

$= x$ otherwise

Lower integral is $0$ and i think that upper integral should be $(b^2 - a^2)/2$(i may be wrong also) but i am not able to prove it.

Answer 1

Since the irrationals are dense, we have $\sup_{x \in [x_{j-1},x_j]}f(x) = x_j$.

Hence, for any partition, $a = x_0 < x_1 < \ldots < x_n = b$, the upper Darboux sum is

$$U(P,f) = \sum_{j=1}^n x_j(x_j - x_{j-1}) = \frac{1}{2} \sum_{j=1}^n(x_j+x_{j-1})(x_j - x_{j-1}) + \frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})(x_j - x_{j-1}) \\ = \frac{1}{2} \sum_{j=1}^n(x_j^2 - x_{j-1}^2) + \frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})^2 = \frac{b^2-a^2}{2} + \frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})^2, $$

Since the last term on the RHS is nonnegative we have, for the upper integral, $$\overline{\int_a^b} f(x) \, dx = \inf_{P}U(P,f) \geqslant \frac{b^2 - a^2}{2}$$

To prove we actually have $\inf_{P}U(P,f) = \frac{b^2 - a^2}{2}$, we show that for any $\epsilon > 0$ there exists a partition such that

$$U(P,f) < \frac{b^2 - a^2}{2} + \epsilon$$

For a uniform partition where $x_j - x_{j-1} = \frac{b-a}{n}$ for $j=1,2, \ldots, n$ we have

$$\frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})^2 = \frac{1}{2} n\frac{(b-a)^2}{n^2} = \frac{(b-a)^2}{2n}$$

Now choose $n > \frac{(b-a)^2}{2\epsilon}$ and we have

$$\begin{align}U(P,f) &=\frac{b^2 - a^2}{2} + \frac{1}{2}\sum_{j=1}^n(x_j-x_{j-1})^2\\ &= \frac{b^2 - a^2}{2}+ \frac{(b-a)^2}{2n} \\&< \frac{b^2 - a^2}{2} + \epsilon\end{align}$$

Therefore,

$$\overline{\int_a^b} f(x) \, dx = \inf_{P}U(P,f) = \frac{b^2 - a^2}{2}$$

Answer 2

For any partition $a=x_0<x_1<\ldots <x_n=b$, the Riemann upper sum is $$ U_{\mathbf x}=\sum_{i=1}^n(x_i-x_{i-1})\sup_{x_{i-1}<x<x_i}f(x) $$ In each interval $(x_{i-1},x_i)$ there is some real numbers $r\in(x_{i-1},x_i)$ and hence the supremum is $\sup_{x_{i-1}<x<x_i}f(x)\ge f(x_{i-1})=x_{i-1}$.

Therefore $$ U_{\mathbf x}\ge \sum_{i=1}^n(x_i-x_{i-1})x_{i-1} $$ Here to compute a bound, we can take any, easy to compute, subdivision. I take $x_i=a+\frac{(b-a)i}{n}=a+ih$

\begin{align} \sum_{i=1}^n(x_i-x_{i-1})x_{i-1}&=h\sum_{i=1}^n(a+(i-1)h) \\ &=\frac{1}{2}h(2a+h(n-1))n \\ &=\frac{1}{2}(b-a)(2a+\frac{(b-a)(n-1)}{n}) \end{align} With $$ \lim_{n\to\infty}\frac{1}{2}(b-a)(2a+\frac{(b-a)(n-1)}{n})=\frac{1}{2}(b^2-a^2) $$ Hence you have your result,

• as $n\to\infty$, the upper bound is $U_x\ge\frac{1}{2}(b-a)^2$
• as upper bound is greater that lower bound, the function is not Riemann integrable