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I have been looking at Kolmogorov's book "Introductory Real Analysis" and have become stuck at the problem 4a) on page 301.

In this problem we are given $f$ a nonnegative and integrable function on $A$, a set of finite measure. We are asked to show that

$\int_A f(x) d \mu \ge 0$

However, how would we prove this? In this text the Lebesgue integral of a function $f$ is defined to be (when the limit exists);

$\int_A f(x) d\mu = \lim\limits_{n \rightarrow \infty} \int_{A} f_n(x) d\mu$

where $f_n$ is a sequence of integrable simple functions that converges uniformly to $f$.

I have attempted to show this by showing that the sequence of simple functions converging to a nonnegative function, is a sequence of nonnegative functions. However this isn't true, take $f_n = -1/n$ and $f = 0$. How else could I go about this?

Thanks.

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  • $\begingroup$ Are you sure that the sets ($A_n$) can also vary? $\endgroup$ – AlexR Apr 3 '14 at 13:07
  • $\begingroup$ Sorry, just a little typo! That's corrected now. $\endgroup$ – Tim Apr 3 '14 at 13:09
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Hint: First show the result for non-negative simple functions. Then use that you can find $f_n \ge 0$ simple with $f_n \nearrow f$ uniformly to see that $$\int_A f d\mu = \lim_{n\to\infty} \underbrace{\int_A f_n d\mu}_{\ge 0} \ge 0$$ For simple functions, the result is fairly easy to prove but depends slightly on the concrete definition of a "simple function" in your book.

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  • $\begingroup$ Is it obvious that there must exist a sequence of simple functions $f_n \ge 0$ such that $f_n \rightarrow f$ uniformly? (The case when $f$ is a simple function is pretty easy, it just follows from the non-negativity of the measure.) $\endgroup$ – Tim Apr 3 '14 at 13:32
  • $\begingroup$ @Tim Well, essentially that's it, yes. $\endgroup$ – AlexR Apr 3 '14 at 13:34
  • $\begingroup$ Thanks for your help. I have got that part sorted now! $\endgroup$ – Tim Apr 3 '14 at 14:00
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Simply state that you are restricting to a sequence simple functions that are non negative.

Or use the fact that $f\ge g\Rightarrow \int f\ge \int g $

Then $\int f = \sup\{\int s|s, simple\}=\sup\{\int s|s, simple,positive\}$

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  • $\begingroup$ Can we always do this? How could we construct such a sequence of functions? $\endgroup$ – Tim Apr 3 '14 at 12:56
  • $\begingroup$ I have put an edit with more justification. $\endgroup$ – Ellya Apr 3 '14 at 12:58
  • $\begingroup$ I have considered using the 'sup' argument, however the definition given of integral is a different one. I know the definitions are equivalent but this property should follow straight from the definition. $\endgroup$ – Tim Apr 3 '14 at 13:00

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