Im currently busy with Measure Theory and noticed that the main theorems that I want to use, require the non-negativity condition.

  1. Fatou's Lemma has the condition that we need a sequence $\{f_n\}$ of non-negative measurable functions.

  2. and Lebesgue Dominated Convergence Theorem a sequence $\{f_n\}$ of measurable functions such that $|f_n| < g$ which will imply $g - |f_n|$ is non-negative.

i) Is it correct that in order to continue with any proof, non-negativity is critical?

ii) According to me, if a sequence is integrable or measurable, it does not satisfy this condition. Is this correct?

iii) I only have $f_n < g$ but it does not mean $g - f_n$ is non-negative.

Apologies, I do not want to state too much as this is part of an assignment. I know how to answer my questions depending on the non-negativity condition.

What might I be missing that could also imply a function/sequence is non-negative?

Thank you!

  • $\begingroup$ I cannot really understand what you are asking in ii) and iii). $\endgroup$ May 27, 2013 at 22:49

1 Answer 1


Some remarks:

  1. Non-negativity is important in part because many authors first define measurability for positive functions (e.g. $L^+$), and then define general measurability using the decomposition $f=f^+-f^-$ of $f$ into positive and negative parts: $$f^+(x)=\max(f(x),0)\quad \text{and} \quad f^-(x)=-\min(f(x),0).$$ Here, we note that $f^+$ and $f^-$ are both positive functions, and we say that $f$ is measurable if both $f^+$ and $f^-$ are (there are issues of indeterminate forms we must avoid, but let's sweep this under the rug). In this sense, non-negativity is important because this is how measures are commonly introduced.
  2. If $f_n <g$, then $g-f_n>0$ (as functions). That is, $g-f_n$ is non-negative. I'm not sure what the confusion is here. I will make clear that the inequalities here are all point-wise, i.e. $g(x)-f_n(x)>0$ for all $x$ in our domain.
  • $\begingroup$ Thank you. Maybe I should rephrase my main question. Does the absolute values not make a difference? Can this be left out in |fn|<g? ie is fn<g only sufficient? $\endgroup$
    – Emilene
    May 28, 2013 at 0:21
  • $\begingroup$ No worries, I figured something out. Thanks for your help $\endgroup$
    – Emilene
    May 28, 2013 at 1:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .