# Property of a convergence series

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).

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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

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
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.