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Suppose I have a function $g \geq 0$ defined by $$g(x) = \int_{-\infty}^{x}f(t)\text{ d}t \geq 0\text{, }x \in \mathbb{R}\text{. }$$

I know for a fact that $g$ is continuous and nondecreasing.

Is this enough to show that $f \geq 0$? If so, I'm not sure how to prove it.

[For those of you familiar with probability, this is trying to show that every PDF $f$ is $\geq 0$ using the necessary and sufficient conditions of the CDF $g$.]

If $f$ is continuous, this is easy. (Use Fundamental Theorem of Calculus, take the derivative, $g$ is nondecreasing, done.)

What if $f$ isn't continuous? I'm stuck.

Edit: As clarified in the comments, I believe that I want to prove that $f \geq 0$ almost surely. However, I am not at all familiar with measure theory besides knowing what a $\sigma$-algebra and measure are (by definition).

Clarinetist
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  • @snarfblaat I am not very familiar with measure theory, but yes, I believe that's correct (that we want $f \geq 0$ a.s.). – Clarinetist Jan 25 '16 at 20:58
  • @snarfblaat Please see my edit for more details – Clarinetist Jan 25 '16 at 21:00
  • http://math.stackexchange.com/questions/580817/can-a-probability-density-function-take-negative-values – Jimmy R. Jan 25 '16 at 21:08
  • @JimmyR. Yes, but I understand now that $f \geq 0$ a.s., i.e., I'm not interested in looking at sets of measure zero for this problem. – Clarinetist Jan 25 '16 at 21:09
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    Ok, assume that $f$ is negative at some set (interval) with positive measure. Then would $g$ still be non-decreasing? – Jimmy R. Jan 25 '16 at 21:10
  • @JimmyR. It's pretty clear to me that my background isn't ready for formally proving this yet. Hopefully I will come back to this question after I've taken some measure theory. Thanks. – Clarinetist Jan 25 '16 at 21:14
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    I think your background is enough. If we denote this interval with $[a,b]$ (where $f<0$) and $λ(a,b)>0$, where $λ$ denotes the Lebesgue measure then we have $g(b)=\int_{-\infty}^b f(t)dt=\int_{-\infty}^af(t)dt+\int_{a}^b f(t)dt<\int_{-\infty}^af(t)dt+λ(a,b)max_{x\in (a,b)}f(x)<\int_{-\infty}^af(t)dt+0=g(a)$ which violates the assumption of nondecreasing. It was necessary in the above calculation that $λ(a,b)>0$. Otherwise there would be no problem. – Jimmy R. Jan 25 '16 at 21:23
  • @JimmyR. The $$\int_{a}^{b}f(t)\text{ d}t < \lambda(a, b)\max_{x \in (a, b)}f(x)$$ part is the only part that's confusing me. I'll look at my measure theory texts later. How is this justified? – Clarinetist Jan 25 '16 at 21:27
  • @JimmyR. Oh, is it the fact that we're working with positive measure? – Clarinetist Jan 25 '16 at 21:31
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    Make it simpler: $$\int_{a}^b f(t)dt < \int_{a}^bmax f(t)dt=\max f(t)\int_{a}^bdt=(b-a)\max f(t)$$ So, if $b-a>0$ and $f(t)<0$ for all $t$ in $(a,b)$ this is negative. – Jimmy R. Jan 25 '16 at 21:31
  • @JimmyR. Clear as a bell. Thank you! Feel free to post that as an answer and I can award points. – Clarinetist Jan 25 '16 at 21:32

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