Let $f$ be integrable over $\mathbb{R}$. Show that the following four assertions are equivalent:
$f = 0$ a.e. on $\mathbb{R}$.
$\int_{\mathbb{R}}fg=0$ for every
bounded measurable function $g$ on $\mathbb{R}$.- $\int_A f = 0$ for
every measurable set $A$.- $\int_{\mathcal{O}} f = 0$ for every open
set $\mathcal{O}$.
$(1) \Rightarrow (2)$
If $f = 0$ a.e. on $\mathbb{R} $, then $m(E_0) = 0$, where $$E_0 =\{x \in \mathbb{R} : f(x) \neq 0\}.$$
Now, by the previous exercise, we shown that for $g: \mathbb{R} \rightarrow \mathbb{R}$ being bounded and measurable, then $fg$ is integrable and
$$\int_{\mathbb{R}}fg = \int_{\mathbb{R}\sim E_0}fg + \int_{E_0}fg,$$
but since $$f = 0 \text{ a.e and } m(E_0) = 0,$$
we have
$$\int_{\mathbb{R}\sim E_0}fg =0,$$
and
$$\int_{E_0}fg = 0,$$
thus,
$$\int_{\mathbb{R}}fg = 0,$$ as required.
$(2) \Rightarrow (3)$
For a measurable set $A$, let $g = \chi_A$. Clearly, $g$ is measurable and bounded. Therefore,
$$0= \int_{\mathbb{R}}fg = \int_{\mathbb{R}}f \chi_A = \int_A f = 0,$$
as required.
$(3) \Rightarrow (4)$
Since every open set is measurable, then if $(3)$ holds, we have
$$\int_{\mathcal{O}}f = 0 \text{ for every open set $\mathcal{O}$, }$$
as required.
$(4) \Rightarrow (1)$
Let $\{U_n\}_{n \in \mathbb{N}}$ be a countable collection of open sets. Then,
$$G = \bigcap_{n \in \mathbb{N}} \mathcal{O}_n,$$
is a $G_\delta$ set with $\mathcal{O}_n = \bigcap_{k \leq n}U_k,$ where each $U_k$ is an open set. Thus, $\{\mathcal{O}_n\}_{n \in \mathbb{N}}$ is a countable descending collection of open sets, i.e.
$$\forall i \in \mathbb{N}: \mathcal{O}_i \supset \mathcal{O}_{i+1}.$$
Now, for $\mathcal{O}_n$ being open, we have
$$\int_{\mathcal{O}_n} f = 0,$$
and by the continuity of integration,
$$\int_{G} f =\int_{\bigcap_{n \in \mathbb{N}} \mathcal{O}_n} f = \lim_{n \to \infty} \int_{\mathcal{O}_n} f = 0.$$
Since every measurable set $E \subset \mathbb{R}$ is of the form $G \sim E_0$, where $G$ is a $G_{\delta}$ subset of $\mathbb{R}$ and $m(E_0) = 0$, so that
$$\int_{E} f = 0 \text{ for each measurable set } E \subset \mathbb{R}.$$
Now, define
$$E^+ = \{x \in \mathbb{R}: f(x) \geq 0 \} \text{ and } E^- = \{x \in \mathbb{R}: f(x) \leq 0 \},$$
then $E^+$ and $E^-$ are measurable subsets of $\mathbb{R}$ and thus,
$$\int_{\mathbb{R}} f^+ = \int_{E^+} f = 0,$$
and
$$\int_{\mathbb{R}} (-f^-) = -\int_{E^-} f = 0,$$
where $f^+ = \max\{f,0\}$ and $f^- = \max\{-f,0\}$. Finally we use the fact (proposition $9$ chapter $4$) that
$$\int_E f = 0 \text{ if and only if } f = 0 \text{ a.e}$$
so that $f^+$ and $f^-$ vanishes almost everywhere in $\mathbb{R}$ and so does $f$ as required.
Best Answer
For (4) $\Rightarrow $ (3), use an argument similar to the one I'm this question (Lebesgue integral on any open set is $\ge 0$, is it still $\geq 0$ on any $G_{\delta}$ set?) to conclude $\int_E f =0$ for every $G_\delta $ set $E $.
Then use that every measurable set $F $ is of the form $F=E\setminus N $ with a null set $N $ and a $G_\delta$ set $E $.
That (3) $\Rightarrow $(1) follows by considering the set $\{x \mid f (x)>0\} $ and $\{x \mid f (x)<0\} $.