# proving $\left|\overline\int_a^b f(x)\ dx\right|\le \overline{\int_a^b}|f(x)|\ dx$

I need to prove that

$$\left|\overline\int_a^b f(x)\ dx\right|\le \overline{\int_a^b}|f(x)|\ dx$$

and show a case that's not valid for $\underline{\int_a^b}$,

using this definition of integrals, that is:

$$\underline{\int_{a}^b}f(x) dx = \sup s(f, P) = \sup \sum_{i=1, t_i \in P} m_i(t_{i}-t_{i-1})$$

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

and

$$\overline{\int_{a}^b}f(x) dx = \inf S(f, P) = \inf \sum_{i=1, t_i \in P} M_i(t_{i}-t_{i-1})$$

where $M_i = \inf\{f(x), x\in [t_i, t_{i-1}]\}$.

I tried to relate things like: $$\sum M_i(t_i-t_{i-1})\le \sum |M_i|(t_i-t_{i-1})$$

I'm considering here that $|\sup A| \le \sup |A|$. If that's true, then I proved what I wanted, right?

The other part of the exercise asks me to prove that an analogus inequality for the $\underline{\int_a^b}$ won't work. I don't know if I'm supposed to prove this:

$$\left|\underline\int_a^b f(x)\ dx\right|> \underline{\int_a^b}|f(x)|\ dx$$ Because no matter what I try, it seems that it's impossible

Consider, for example,

$$f(x)=\begin{cases} -1 \mbox{ if x\in [-1,0)} \\ +1 \mbox{ if x\in [0,1]} \end{cases}$$

Let $P=\left \{ a,x_1,\cdots, x_{n-1},b \right \}$ be a partition of $[a,b]$;

$M_i=\sup \left \{ f(x):x_{i-1}\le x\le x_i \right \}$; and

$N_i=\sup \left \{ \vert f(x)\vert:x_{i-1}\le x\le x_i \right \}$.

Note that $\vert M_i\vert \le N_i$.

Then,

$\tag1\left|\overline\int_a^b f(x)\ dx\right|\le \left | \sum_{i=1}^{n}M_i(x_i-x_{i-1}) \right |\le \sum_{i=1}^{n}\vert M_i\vert (x_i-x_{i-1})\le \sum_{i=1}^{n}N_1 (x_i-x_{i-1})$.

Thus,

$\tag2\left|\overline\int_a^b f(x)\ dx\right|\le \sum_{i=1}^{n}N_1 (x_i-x_{i-1})$.

But this is true for all partitions, so if we inf over the $RHS$ we get

$\tag3 \left|\overline\int_a^b f(x)\ dx\right|\le \overline{\int_a^b}|f(x)|\ dx$

which is what we want.