# Riemann integrable functions: $\left|\int_a^bf(x)\,\mathrm dx\right|\le\int_a^b|f(x)|\,\mathrm dx$

Suppose that $f\colon [a,b]\rightarrow \mathbb R$ is bounded and has finitely many discontinuities. Show that as a function of $x$ the expression $|f(x)|$ is bounded with finitely many discontinuities and is thus Riemann integrable. Then show that: $$\left|\int_a^bf(x)\,\mathrm dx\right|\le\int_a^b|f(x)|\,\mathrm dx.$$ I have no idea where to even begin this problem.

• If $f$ is bounded then there exists some $M > 0$ such that $-M < f(x) < M$ for all $x\in[a,b]$. Can you see how this implies that $|f|$ is bounded? I'm not sure why you were downvoted. This is a perfectly reasonable question.. – Cameron Williams Jul 19 '13 at 15:19
• Let $v(x)=|x|$. Then $v(x)$ is a continuous function, so if $f(x)$ is continuous at $c$, then $v(f(x))$ is continuous at $c$. The second part needs some work. If you have no theorems, one can go back to the definition of Riemman integral. We can use the Triangle Inequality on the Riemann sums. – André Nicolas Jul 19 '13 at 15:23
• @CameronWilliams it implies |f| is bounded because it would then turn into |f| $\le$ $M$ correct? – user72195 Jul 19 '13 at 15:51
• @user72195 That is exactly correct. – Cameron Williams Jul 19 '13 at 15:54
• @CameronWilliams so then how would I go on to show the next inequality? – user72195 Jul 19 '13 at 15:56

hint: $|f(x)| = \max(f(x), 0) - \min(f(x), 0)$

hint: $f(x) = \max(f(x), 0) + \min(f(x), 0)$

hint: if f and g are Riemann integrable functions, then so is $f+g$, $\int (f+g) = \int f + \int g$

First, note$^{\dagger}$ that $$||f(x)|-|f(a)||\leq |f(x)-f(a)|$$

so that $|f|$ as at most the same number of discontinuities $f$ has. Reason: if $f$ is continuous at $x=a$, the above inequality implies $|f|$ is continuous too.

$\dagger$: This is a "reversed" version of the triangle inequality. The usual one is $$|x'+y|\leq |x'|+|y|$$

Now, put $x'=x-y$. Then $|x|\leq |x-y|+|y|$ and thus $|y|\leq |x-y|+|x|$. This gives $-|x-y|\leq |x|-|y|\leq |x-y|$ which is $||x|-|y||\leq |x-y|$, the "reverse" triangle inequality.

Recall that a (tagged) Riemann sum for $f$ in $[a,b]$ looks like $$S(f,P)=\sum_{i=1}^n f(t_i)(x_i-x_{i-1})$$

A tagged partition for $|f|$ will have the form $$S(|f|,P)=\sum_{k=1}^n |f(t_i)|(x_i-x_{i-1})$$

Note that \begin{align}|S(f,P)|&=\left|\sum_{i=1}^nf(t_i)(x_i-x_{i-1})\right|\\ &\leq \sum_{i=1}^n |f(x_i)||x_i-x_{i-1}|\\&=\sum_{i=1}^n |f(x_i)|(x_i-x_{i-1})=S(|f|,P)\end{align}

For we always assume $x_i>x_{i-1}$. Since we assume $f$ is bounded, this means that there is a constant $M$ such that $$-M\leq f(x)\leq M$$

But we can write this as $$|f(x)|\leq M$$

Since the absolute value is always positive, we get $$0\leq |f(x)|\leq M$$

As $f$ is bounded, $\exists M > 0$ such that $-M \leq f(x) \leq M$ for all $x \in [a,b]$. This implies that $|f(x)| \leq M$ for all $x \in [a,b]$. Thus, $|f|$ is bounded. Again as absolute values is a continuous function, $|f|$ is continuous wherever $f$ is continuous. Hence $|f|$ has at most finitely many discontinuities. Now, we know that a function with finitely many discontinuities is Riemann integrable.

For the inequality, choose $c = \pm 1$ such that $c\int f > 0$. Then

$$\left|\int f\right| = c\int f = \int cf \leq \int |f|$$

since $cf \leq |f|$.