4

Claim: If f is integrable, $\left|\int_a^bf(x)dx\right|\le\int_a^b|f(x)|dx$

Proof (attempt):

We know $-|f|\le f \le|f|$,

so $\int-|f| \le \int f \le \int|f|$.*

Since, if $-b<a<b$, we say $|a|<b$,

we can say $|\int f| \le \int |f|$

*: Is this true? If so, how can I prove this?

Would this be a rigorous enough proof (well, I did neglect to write out a,b for the integrals, but besides that)? It just seems too simple for some reason.

Martin Sleziak
  • 53,687
MathMathCookie
  • 2,156
  • This follows from the fact that if $f\leq g$ integrables in $[a,b]$ then $\int_a^b f\leq \int_a^b g$. All that you did is fine, but you need to prove that if $f$ is integrable then $|f|$ is integrable. This allows you to do all the work with the inequalities, that you actually did it. – leo Nov 22 '11 at 05:38
  • @leo As far as I know the definition for integrability is that $f$ is integrable if $\int |f|$ is finite. – Ashok Nov 22 '11 at 05:54
  • @Ashok: yep, for Lebesgue integral – leo Nov 22 '11 at 06:02

3 Answers3

7

You have the right ideas, but you worry too much about $a$ and $b$. The correct way to state this proof would be to say this :

Since $$-|f(x)| \le f(x) \le |f(x)|,$$ you have $$ - \int_a^b |f(x)| \, dx = \int_a^b -|f(x)| \, dx \le \int_a^b f(x) \, dx \le \int_a^b |f(x)| \, dx$$ and therefore $$ \left| \int_a^b f(x) \, dx \right| \le \int_a^b |f(x)| \, dx. $$

P.S. : I am using the fact that $f \le |f|$ implies $\int f \le \int|f|$, and the linearity of the integral when I pull out the minus. I don't know if you are working with the Riemann or the Lebesgue integral, but it both cases it is usually proven during the course.

Hope that helps,

EDIT : I understand that your problem was to find out why if $f$ was integrable then so was $|f|$. If you use the Lebesgue integral, saying that $f$ is integrable is an exact synonym of saying that $|f|$ is since $|f| = \|f\|$. If you are working with the Riemann integral, note that $g(x) = |x|$ is a continuous function, and it is shown that the composition of a continuous function and an integrable one remains integrable in many analysis texts. Perhaps you should use this.

Srivatsan
  • 26,311
Patrick Da Silva
  • 41,413
  • That's fine @Patrick Da Silva, but my point is: why is trivial that the existence of $\int_a^b f$ implies the existence of $\int_a^b |f|$? This don't follow immediately. I think that part of the objective of this exercise is discover why this is true – leo Nov 22 '11 at 05:54
  • 1
    Oh, so that's why you were worrying. Are you working with Riemann or Lebesgue integral? – Patrick Da Silva Nov 22 '11 at 05:56
  • With Lebesgue integral is immediately. With the Riemann integral there is something more to do. I know how to prove it. Just, I want that OP think about it. – leo Nov 22 '11 at 06:01
  • There is not really much to do. If OP was worried about such a thing it's probably because he would've not found it himself if he had searched for a long time, or he was lazy. In both cases I didn't mind answering myself since I knew. Sorry leo. – Patrick Da Silva Nov 22 '11 at 06:12
  • Now is better I think. – leo Nov 22 '11 at 06:40
  • @PatrickDaSilva I appreciate your edit though I did know this fact beforehand and proved it differently using partitions. I hadn't thought to prove it using the fact that the composition of a continuous function and an integrable one remains integrable, so thank you. I was just worried that I had missed something b/c the proof seemed so simple, when the result seemed pretty powerful IMO. – MathMathCookie Nov 22 '11 at 22:52
  • @MathMathCookie : No problem! Keep asking, we keep answering. =) – Patrick Da Silva Nov 24 '11 at 04:15
  • Hello! Can you please explain why $| \int f(x)dx | \le \int |f(x)|dx$? I didn't understand your last step there, why is it true. Thanks! –  Apr 23 '21 at 13:34
  • @TheCalc: It has nothing to do with integrals, it's a general fact about numbers. The inequality $|A| \le B$ is equivalent to $-B \le A \le B$. Take $A = \int_a^b f(x) , dx$ and $B = \int_a^b |f(x)| , dx$. – Patrick Da Silva Apr 24 '21 at 01:45
1

Your proof looks OK, but a few comments...

(1) It is a fact that if $g(x) \leq f(x)$ for all $x \in [a,b]$ then $\int_a^b g(x) dx \leq \int_a^b f(x) dx$, and your inequality (*) follows from that. If you've proven that fact in class (and I'm guessing you have) then it's OK to use it in your proof.

(2) To hit the final step in your proof, you technically will want $\int - |f| = - \int |f|$. That's also true, and I'm guessing you've proven that in class too.

Dimitrije Kostic
  • 1,550
1

With the Lebesgue's theory, your claim is just a comment that follows a sequence of observations after the definition of integrable function.

I recall this exercise from early courses. So, I think that the OP talks about the Riemann integral. As I said in the comments, all the reasoning with the inequalities is correct provided that we are working with Riemann integrable functions. Let me see why this is the case.

Theorem If $f$ is Riemann integrable in $[a,b]$, then $|f|$ is Riemann integrable in $[a,b]$.

Proof. Let $P=\{a=x_0,\ldots x_n=b\}$ a partition of $[a,b]$. For each $k\in \{1,\ldots,n\}$ let $M_k(f)=\sup f([x_{k-1},x_k])$ and $m_k(f)=\inf f([x_{k-1},x_k])$. Then we have $$\begin{align*} M_k(|f|)-m_k(|f|)&=\sup \{|f(x)|-|f(y)|:x,y\in [x_{k-1},x_k]\}\tag{1}\\ &\leq \sup \{|f(x)-f(y)|:x,y\in [x_{k-1},x_k]\}\\ &= M_k(f)-m_k(f), \end{align*}$$ multiplying both sides by $x_k-x_{k-1}$ and summing over $k$ we get $$0\leq U(P,|f|)-L(P,|f|)\leq U(P,f)-L(P,f).$$ Since $f$ is integrable, we can do the difference between the upper and lower sums as small as we want by taking the correct partitions. The above inequality shows that we can do the same with the upper and lower sums of $|f|$, and that implies that $|f|$ is integrable in $[a,b]$.

To see why $(1)$ holds, pick $k\in\{1,\ldots,n\}$. For each $x,\ y\in[x_{k-1},x_k]$, we know that $$M_k(|f|)\geq |f(x)|\qquad\text{and}\qquad -m_k(|f|)\geq -|f(y)|.$$ So $$M_k(|f|) - m_k(|f|)\geq |f(x)| - |f(y)|\qquad \forall x,y\in [x_{k-1},x_k],$$ $$M_k(|f|) - m_k(|f|)\geq \sup\{|f(x)| - |f(y)| : x,y\in [x_{k-1},x_k]\}.$$

In the other hand, by the properties of the sup, for each $j\in\mathbb N$, there are $x_j\ y_j\in [x_{k-1},x_k]$ such that $$\sup\{|f(x)| - |f(y)| : x,y\in [x_{k-1},x_k]\} - \frac 1j \leq |f(x_j)| - |f(y_j)|,$$ but $|f(x_j)| - |f(y_j)|\leq M_k(|f|) - m_k(|f|)$ for each $j$ as we saw above, so we get $$\sup\{|f(x)| - |f(y)| : x,y\in [x_{k-1},x_k]\} - \frac 1j \leq M_k(|f|) - m_k(|f|)\qquad \forall j\in\mathbb N,$$ thus \begin{align*} \lim_{j\to\infty} \left(\sup\{|f(x)| - |f(y)| : x,y\in [x_{k-1},x_k]\} - \frac 1j\right) &\leq M_k(|f|) - m_k(|f|)\\ \sup\{|f(x)| - |f(y)| : x,y\in [x_{k-1},x_k]\} &\leq M_k(|f|) - m_k(|f|). \end{align*} Therefore, $(1)$ holds for each $k\in\{1,\ldots,n\}$.

leo
  • 10,433