# Prove $|\int_a^b$$f(x)dx| \leq \int_a^b$$|f(x)|dx$

Prove $$\left|\int_a^b f(x)dx\right| \leq \int_a^b |f(x)|dx.$$

My thoughts: first I think we must show that if $f \geq 0$ is Riemann integrable on $[a,b]$, then $\int_a^b f(x)dx \geq 0$. Then we can show that if $f$ and $g$ are integrable on $[a,b]$ and $f \leq g$, $\int_a^b f(x)dx \leq \int_a^b g(x)dx$. Then we can use these facts to prove the above.

Some facts that are useful:

Riemann's Condition: Suppose $f:[a,b] \to R$ is bounded. Then $f$ is Riemann integrable if and only if for each $\epsilon > 0$, there is a partition $P$ of $[a,b]$ s.t $U(f,P) - L(f,P) < \epsilon$

Also, $f$ is Riemann integrable if $\sup_P L(f,P) = inf_P U(f,P)$. Thus $\int_a^b f(x)dx$ is the common value.

Partition of $[a,b]$ is $P=\{a=x_0 < x_1 < \ldots < x_{n-1} < x_n\}$
$\Delta_j = x_j - x_{j-1}$
$\mathrm{mesh}(P)= \max\{\Delta_j : 1 \leq j \leq n\}$

$U(f,P) = \sum_{j=1}^n \sup\{f(x): x_{j-1} \leq x \leq x_j\} * \Delta_j$
$L(f,P) = \sum_{j=1}^n \inf\{f(x): x_{j-1} \leq x \leq x_j\} * \Delta_j$

Lemma: If $P$ and $Q$ are partitions of $[a,b]$, then $L(f,P) \leq U(f,Q)$

Theorem: $f$ bounded on $[a,b]$ is Riemann integrable if and only if:

For every $\epsilon > 0$, there is $\delta > 0$ such that every partition $Q$ with $\mathrm{mesh}(Q) < \delta$ satisfies $U(f,Q) - L(f,Q) < \epsilon$

• I thought it follows directly from triangle inequality and the definition of Riemann integrals. – Divide1918 Nov 6 at 14:50

Hint: If $f,g$ are Riemann integrable and bounded functions then $\max \{f,g\},\min \{f,g\}$ is also Riemann integrable.

As $f$ is Riemann integrable $\forall \epsilon >0\exists f_1,f_2 :f_1\le f\le f_2$ piecewise constant functions such that $$\int f_2-\epsilon\le \int f\le \int f_1+\epsilon$$

As $g$ is Riemann integrable $\forall \epsilon >0\exists g_1,g_2 :g_1\le g\le g_2$ piecewise constant functions such that $$\int g_2-\epsilon\le \int g\le \int g_1+\epsilon$$

We have that $$f_2= f_2+f_1-f_1\le f_2-f_1+f_1+g_2-g_1=(f_2-f_1+g_2-g_1)+f_1$$

Similarly $g_2\le (f_2-f_1+g_2-g_1)+f_1$ which implies

$$\max\{f_2,g_2\}\le (f_2-f_1+g_2-g_1)+\max\{f_1,g_1\}$$

that is

$$\max\{f_2,g_2\}-\max\{f_1,g_1\}\le (f_2-f_1+g_2-g_1)$$

so

\begin{align}\int_U\max\{f,g\}-\int_L\max\{f,g\}&\le&\\ \int \max\{f_2,g_2\}-\int \max\{f_1,g_1\}& \le& \\ \int(f_2-f_1+g_2-g_1)\le 2\epsilon\end{align}

And write $|f|=\max\{f,0\}-\min\{f,0\}$.

So $|f|$ is also Riemann integrable.

Now use,

$$-|f|\le f\le |f|$$ $$\Rightarrow -\int_A|f|\le \int_{A}f\le \int_{A}|f|$$ $$\Rightarrow |\int_A f|\le \int_A |f|$$ where $A=[a,b]$

• I proved $|f|$ is integrable assuming that you wanted it to be proved(as you wrote the useful facts) – Abhra Abir Kundu Apr 3 '13 at 16:05
• thanks, but i still dont understand the proof |f| is integrable. Could you put more detail, thanks – sarah Apr 3 '13 at 16:07
• @maria now i think its clear.... – Abhra Abir Kundu Apr 3 '13 at 16:40
• @AbhraAbirKundu Sorry, I read $\int \max\{f_2,g_2\}$ as $\int \max\{f,g\}$! I deleted the comment. Great answer. – Pedro Tamaroff Apr 3 '13 at 19:37

Try to prove that $U(|f|,P) - L(|f|,P) \le U(f,P) - L(f,P)$. Once you do this, the fact that $|f|$ is Riemann integrable whenever $f$ is Riemann integrable is immediate.