Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I stumbled onto this exercise while studying for an exam. I thought it looked fun.

Let $f$ be integrable over $\mathbb{R}$. Show the following four assertions are equivalent.

(i) $f=0$ a.e. on $\mathbb{R}.$

(ii) $\int_\mathbb{R} fg=0$ for every bounded measurable function $g$ on $\mathbb{R}.$

(iii) $\int_A f =0$ for every measurable set $A$.

(iv) $\int_O f = 0$ for every open set $O$.

Just a few observations.

(i) implies (ii) Does this follow from how the function is defined and this inequality: $\int_\mathbb{R} fg \leq \int_\mathbb{R} f\cdot M =0$.

(iii) implies (iv) does that work out because for $\int_E f = 0$ iff $f=0$ a.e. on $E$ and the fact that open sets are also measurable sets.

I don't actually want a rigorous proof of this. I'm just interested in fine tuning my intuition about integrable functions and pulling together definitions.

Edit: I do not have a strong intution of going from (ii) to (iii).

share|cite|improve this question
Note that the characteristic function of $A$ is a measurable function and that it is certainly bounded. Then look up the definition of $\int_A f$ for $A \subset \mathbb{R}$. – Adam Saltz Nov 27 '12 at 3:41
$(ii)\Longrightarrow (iii)\,$ : choose $\,g(x)=1\,$ , then what (ii) says is simply that $\,\int_{\Bbb R}f=0\,$ ... – DonAntonio Nov 27 '12 at 10:59
up vote 1 down vote accepted

I didn't work out the rigor, so this may not all be right, but it seems clear that $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4)$, since if $f=0$ a. e., $fg=0$ a. e.; we can let $g(x)=1$ if $x\in A$ and let $g=0$ otherwise; and all open sets can be written as unions/intersections of open intervals, which are clearly measurable, so $\{X|X$ is open$\}\subset\{X|X$ is measurable$\}$. We also have an intuition for $(3)$ and $(4)$ implying $(1)$, since we know that if $\int_{A}{fdx}=0$ and $f\neq 0$ we should be able to break $A$ up into a possibly infinite number of open subsets each of which have $\int_{A}fdx\neq 0$, but we know this should be zero by the assertions given in $(3)$ and $(4)$ (for $(3)$, just replace "open" with measurable -- it should be the same). Then, we see that $(2)\Rightarrow (3)\Rightarrow (1)$, so $(2)\Rightarrow (1)$. From there, we see they all imply each other from the simple logic. For your intuition's sake, $(2)$ implies $(1)$ since we can let $g$ be the same function as used to imply $(3)$, and we can use the same reasoning as in $(3)$ and $(4)$. Just by the logic we merely needed to show $(1)\Rightarrow (2)\Rightarrow (3)\Rightarrow (4)\Rightarrow (1)$. Hope this helped you and your intuition

share|cite|improve this answer

Well, I think it's simpler to observe that (i)$\Longrightarrow$(ii) because $\,f=0\,\, a.e.\Longrightarrow fg=0\,\,a.e.$. The condition on $\,g\,$ is only to be sure $\,fg\,$ is integrable.

share|cite|improve this answer
Bounded measurable functions are indeed integrable. – emka Nov 27 '12 at 3:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.