Need explanation of passage about Lebesgue/Bochner space From a book:

Let $V$ be Banach and $g \in L^2(0,T;V')$. For every $v \in V$, it holds that
  $$\langle g(t), v \rangle_{V',V} = 0\tag{1}$$
  for almost every $t \in [0,T]$.

What I don't understand is the following:

This is equivalent to $$\langle g(t), v(t) \rangle_{V',V} = 0\tag{2}$$ for all $v \in L^2(0,T;V)$ and for almost every $t \in [0,T]$.

OK, so if $v \in L^2(0,T;V)$, $v(t) \in V$, so (2) follows from (1). How about the reverse? Also is my reasoning really right? I am worried about the "for almost every $t$ part of these statements, it confuses me whether I am thinking correctly.
Edit for the bounty: as Tomas' comment below, is the null set where (1) and (2) are non-zero the same for every $v$? If not, is this a problem? More details would be appreciated.

Your reasoning for (1) implies (2) is almost correct: $v\in L^2(0,T;V)$ implies $v(t)\in V$ only for almost every $t$. The function $v(t) = (t-a)^{-1/3} \in L^2(0,T;\mathbb R)$ for some number $a\in [0,T]$ serves as an example. Consequently, for a given $v\in L^2(0,T;V)$, we also have to exclude the points where $v(t)\not\in V$.
Nevertheless, (2) requires the statement for all $v$, such that the remark above is just a technicality.
