Let $f, g : [a,b] \to \mathbb R$ be two continuous functions such that $g(x) \le f(x)$ for all $x \in [a, b]$

Question

Let $f, g : [a,b] \to \mathbb R$ be two continuous functions such that $g(x) \le f(x)$ for all $x \in [a, b]$. Define
$$\phi(x) =
\begin{cases}
f(x), & x \in \mathbb Q \cap [a,b] \\
g(x), & x \in [a,b] \setminus \mathbb Q
\end{cases}$$
show that
$$\overline{\int_a^b} \phi = \int_a^b f \\{\text{and}}\\ \underline{\int_a^b} \phi = \int_a^b g $$

What have i managed to do so far :

Case 1: $g(x)=f(x)$, then $$\overline{\int_a^b} \phi = \int_a^b f = \underline{\int_a^b} \phi = \int_a^b g $$

Case 2: $g(x) \lt f(x)$, let $P: a=t_0 \lt … \lt t_n = b$

$ m_i = inf f\{ f(x): x \in [t_{i-1},t_i]\}$

$M_i = sup g\{ g(x): x \in [t_{i-1},t_i]\}$

then

$S(g,P) = \sum_{i=1}^n M_i (t_i – t_{i-1}) = sup g \sum_{i=1}^n (t_i – t_{i-1}) = sup g (b-a) $

$s(f,P) = \sum_{i=1}^n m_i (t_i – t_{i-1}) = inf f \sum_{i=1}^n (t_i – t_{i-1}) = inf f (b-a)$

$\overline{\int_a^b} \phi = inf f (b-a)$

$\underline{\int_a^b} \phi = sup g (b-a)$

Thanks in advance for any help.

Answer 1

For any partition $P=[t_0,t_1,\dots,t_n]$ of $[a,b]$, because $\mathbb{Q}$ is a dense subset of $\mathbb{R}$, you can choose $s_1,s_2,\dots,s_n \in \mathbb{Q}$ such that for all $k$ $$s_k \in [t_k,t_{k-1}]$$ So according to definition of upper Riemann integral $$\overline{\int_{a}^b} \phi \ge \sum_{k=1}^n \phi(s_k)( t_k-t_{k-1})\stackrel{ s_i \in \mathbb{Q}}{=}\sum_{k=1}^n f(s_k)( t_k-t_{k-1})$$ for all choice of $P$ and $(s_i)$.
Besides, because $f$ is continuous hence it is Riemann integral on $[a,b]$. Hence as the norm(the mesh) of $P$ converges to $0$ $$\sum_{k=1}^n f(s_k)( t_k-t_{k-1}) \longrightarrow \int_{a}^b f$$ Thus, $$\overline{\int_{a}^b} \phi \ge \int_{a}^b f$$ On the other hand, $\phi \le f$, hence $$\overline{\int_{a}^b}\phi \le \overline{\int_{a}^b } f = \int_{a}^b f$$ Thus $$\overline{\int_{a}^b} \phi = \int_{a}^b f$$

You can argue similarly with the other identity.