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Let $f:[a,b]\to\mathbb{R}$ be Lebesgue integrable.

Clearly, if $f$ is non-negative then $$ g:[a,b]\ni x\mapsto\int_a^x f(t)\,\mathrm{d}t\in\mathbb{R} $$ is non-decreasing since for $x<y$ it holds $$ g(y)-g(x) =\int_a^y f(t)\,\mathrm{d}t-\int_a^x f(t)\,\mathrm{d}t =\int_x^y f(t)\,\mathrm{d}t\ge 0\text{.} $$

How can I show that $f$ is non-negative almost everywhere if $g$ is non-decreasing?

I would suppose that $f$ is negative on a set $A\subseteq[a,b]$ of positive measure. Then there exists a compact set $B\subseteq A$ of positive measure and it holds $$ \int_B f(t)\,\mathrm{d}x<0\text{.} $$ Does $B$ contain a non-empty interval?

precarious
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2 Answers2

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I would use the monotone class theorem. Assume without loss of generality that $f$ is Borel measurable (since there is a Borel measurable function $\tilde{f}$ with $f = \tilde{f}$ a.e.).
Let $\mathcal{M}$ be the collection of all Borel subsets $E \subset [a,b]$ for which $\int_E f \ge 0$. Then $\mathcal{M}$ is a monotone class (use the dominated convergence theorem). Let $\mathcal{A}$ be the collection of all finite unions of half-open intervals $[c,d)$. Then $\mathcal{A}$ is an algebra, $\mathcal{A} \subset \mathcal{M}$, and $\mathcal{A}$ generates the Borel $\sigma$-algebra. By the monotone class theorem, $\mathcal{M}$ equals the Borel $\sigma$-algebra. In particular, $\{f < 0\} \in \mathcal{M}$, from which it follows that $f \ge 0$ a.e.

See also this related question.

Glorfindel
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Nate Eldredge
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  • Thank you for your answer. In spite of your very helpful references I do not understand every step. Unfortunately, my knowledge of measure theory is too limited. Does every $S\in\mathcal{M}$ of positive measure contain a non-empty interval? Otherwise, I feel completely lost with your last sentence. :-( – precarious Jan 14 '12 at 22:20
  • @precarious: No, not every $S \in \mathcal{M}$ of positive measure will contain an interval. (Indeed, $\mathcal{M}$ turns out to contain every Borel set, and there are certainly Borel sets of positive measure containing no interval, as Brian Scott's comment indicates.) In the last sentence, ${f < 0}$ is shorthand for ${x : f(x) < 0}$, and I am pointing out that the integral of $f$ over this set is nonnegative. Since $f$ is strictly negative on this set, it must be that the set has measure zero. – Nate Eldredge Jan 14 '12 at 22:32
  • @precarious: I've intentionally left some of the details in my answer for you to work out; it will be educational. – Nate Eldredge Jan 14 '12 at 22:32
  • Oh yes, now I understand your argument. Thank you! I am convinced that it is a good idea for me to thoroughly work through measure theory including this example. For the moment, I am much deeper into it than I originally wanted to go. However, I am going to return when my schedule admits. – precarious Jan 14 '12 at 22:49
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Inspired by Jonas' answer to my related question

I did some research and can now come up with an alternative solution.

In the book The Integrals of Lebesgue, Denjoy, Perron, and Henstock by Russell A. Gordon, it is stated that:

A non-decreasing function $g:[a,b]\to\mathbb{R}$ is differentiable almost everywhere and $g'$ is non-negative where it exists.

Since $f=g'$ almost everywhere, $f$ is non-negative almost everywhere.

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precarious
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