If f is integrable over E iff |f| is integrable over E.

[futurebird]

Homework Statement

Royden Chapter 4, Problem 10a

Show that if f is integrable over E then so is |f| and \left|\int_E f \right| \leq \int |f|.
Does the integrability of |f| => the integrability of f?

Homework Equations

f^+ = max\{f, 0\}
f^- = max\{-f, 0\}

|f| = f^+ + f^-

A function is integrable if \int f < \infty

The Attempt at a Solution

I have \left|\int_E f \right| \leq \int |f|

\left|\int_E f\right| = \left| \int_E f^+ - \int_E f^- \right| \leq \left| \int_E f^+ \right| + \left| \int_E f^- \right| = \int_E |f|.

Next I want to show that if f is integrable over E then so is |f|.

\int_E f < \infty

\int_E f^+ - \int_E f^- < \infty

f+ and f- are finite because the expression a-b is only finite if both a and b are finite.

\int_E f^+ < \infty
\int_E f^- < \infty

Hence:

\int_E |f| = \int_E f^+ + \int_E f^- < \infty.


Does the integrability of |f| => the integrability of f?

I would say yes for similar reasons. I don't feel very confident about my answers here. Is the reasoning correct?

Also part b says that the improper Riemann integral (a limit) may exists for a function when the Legesgue integral fails to exists. and gives (sin x)/x as an example. Is the problem with (sin x)/x that f+ and f- are not finite??

 

[rochfor1]

To see that |f| is integrable => f is integrable, think about the inequality you proved.

 

[JSuarez]

The proof of the inequality

<br /> \left|\int_E f \right| \leq \int |f|<br />

is correct if you assume that f is integrable.

f+ and f- are finite because the expression a-b is only finite if both a and b are finite.

And what if both are infinite? Does this contradict the hypothesis that f is integrable?

By the way, a function is Lebesgue integrable iff is absolutely Lebesgue integrable (that is f is integrable iff |f| also is).

The reason regarding the existence of improper Riemann integrals is pretty much what you said, but note that the how problem ties with your quoted statement above.

 

[futurebird]

The proof of the inequality

<br /> \left|\int_E f \right| \leq \int |f|<br />

is correct if you assume that f is integrable.

And that was given, so it's just very simple.

And what if both are infinite? Does this contradict the hypothesis that f is integrable?

Yes.

 

[JSuarez]

So you have it.

 

[r.a.c.]

But I don't get it. Graphically that is. The integral on R of higher dimensions is analogue to summation of R. Now if f = sinx /x the summation (if x belongs to N) is finite but |sinx/x| is not finite. Doesn't the same apply for the integral? Let x belong to R of higher dimensions. Wouldn't the same happen? Because I have never seen what you said. I have seen and tried to prove what you are asking but never what you said.

 

[JSuarez]

Sorry, you have to more clear. I don't understand your reply.