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Let $f,g:[0,1]\to\mathbb{R}$ be defined by

$ f(x)= \begin{cases} 0& \text{ if }x\in\mathbb{Q},\\ -1& \text{ if }x\notin\mathbb{Q},\\ \end{cases}$

and

$g(x)= \begin{cases} -1& \text{ if }x\in\mathbb{Q},\\ 0& \text{ if }x\notin\mathbb{Q},\\ \end{cases}$

(a) Compute the upper integrals $\displaystyle\overline{\int_{0}^{1}}f(x)dx$ and $\displaystyle\overline{\int_{0}^{1}}g(x)dx$.

(b) Is it true that $\displaystyle\overline{\int_{0}^{1}}(f(x)+g(x))dx = \displaystyle\overline{\int_{0}^{1}}f(x)dx+\displaystyle\overline{\int_{0}^{1}}g(x)dx $?

Here are my solutions, I am wondering if they make sense.

Solution for part (a):

$S_g(D) = \sum_{i=1}^{n} \delta_i G_i \text{ where } G_i = sup\{g(x): x_{i-1} < x < x_i\}$ $S_f(D) = \sum_{i=1}^{n} \delta_i F_i \text{ where } F_i = sup\{f(x): x_{i-1} < x < x_i\}$

On any interval, there are irrational and rational numbers. This means that:

$sup\{g(x): x_{i-1} < x < x_i\} = sup\{f(x): x_{i-1} < x < x_i\} = 0$

As a result, for all divisons $D$ of [0 ,1]:

$\sum_{i=1}^{n} \delta_i G_i = \sum_{i=1}^{n} \delta_i (0) = 0$

$\sum_{i=1}^{n} \delta_i F_i = \sum_{i=1}^{n} \delta_i (0) = 0$

Since the $S_g(D) = S_f(D) = 0$ for all divisions of $[0, 1]$, it follows that:

$inf\{S_f(D): \text{D is a division of [0, 1}]\} = inf\{S_g(D): \text{D is a division of [0, 1]}\} = 0$

Therefore $\displaystyle\overline{\int_{0}^{1}}f(x)dx = \displaystyle\overline{\int_{0}^{1}}g(x)dx = 0$

Solution for part (b):

This is not true. Consider the functions $f$ and $g$ from part a). We showed that

$\displaystyle\overline{\int_{0}^{1}}f(x)dx = \displaystyle\overline{\int_{0}^{1}}g(x)dx = 0$.

So $\displaystyle\overline{\int_{0}^{1}}f(x)dx + \displaystyle\overline{\int_{0}^{1}}g(x)dx = 0$

let $h(x) = g(x) + f(x)$.

if $x \in \mathbb{Q}$, we have that $f(x) = 0$ and $g(x) = -1 \implies h(x) = -1 + 0 = -1 \text{ for } x \in \mathbb{Q}$

if $x \not\in \mathbb{Q}$, we have that $f(x) = -1$ and $g(x) = 0 \implies h(x) = 0 + -1 = -1 \text{ for } x \not\in \mathbb{Q}$

it follows that $h(x) = -1$ and therefore the $sup\{ h(x): x_{i-1 } < x < x_i\} = -1.$

$S_h(D) = \sum_{i=1}^{n} \delta_i H_i $ where $H_i = sup\{ h(x): x_{i-1 } < x < x_i\}$ $= \sum_{i=1}^{n} \delta_i (-1) = \sum_{i=1}^{n} -\delta_i$

since we are on the interval $[0, 1]$, $\sum_{i=1}^{n}\delta_i = 1 \implies \sum_{i=1}^{n} -\delta_i = -1$

Therefore, the upper Darboux sums of $h(x)$ for any division $D$ of $[0, 1] =-1$.

This means that $inf\{S_h(D): \text{ D is a division of [0, 1]}\} = -1$

and $\displaystyle\overline{\int_{0}^{1}}h(x)dx = \displaystyle\overline{\int_{0}^{1}}(f(x) + g(x))dx = -1 \ne \displaystyle\overline{\int_{0}^{1}}f(x)dx + \displaystyle\overline{\int_{0}^{1}}g(x)dx = 0$

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Yes, it is correct. However, concerning part b), after having observed that $h=-1$, I would then simply add that $-1$ is Riemann-integrable and that$$\overline{\int_0^1}h(x)\,\mathrm dx=\int_0^1-1\,\mathrm dx=-1\times(1-0)=-1.$$

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