Showing that a function is a weak time derivative of a Bochner function

Question

This question is a follow-up of If $u$ is a weak solution of the Navier-Stokes equations then $\partial_t u \in L^{4/3}(0,T; V^*)$ which I've split into multiple posts to help keep things focused.

Let $V \subset H^1$ and $H \subset L^2$ denote function spaces and $u \in L^\infty(0,T; H) \cap L^2(0,T; V)$. Using the following lemma from the book I am following:

Lemma 3.6 If $u$ is a weak solution of the Navier-Stokes equations then $$\int_{t_1}^{t_2} \langle \nabla u, \nabla \phi\rangle + \int_{t_1}^{t_2} \langle(u \cdot \nabla )u, \phi \rangle = \langle u(t_1), \phi \rangle - \langle u(t_2), \phi \rangle \tag{3.4}$$ for $\phi \in C_{c, \sigma}^\infty(\Omega)$ and almost every $t_2 \geq t_1$, for almost every $t_1 \geq 0$ including $t=0$.

for any compactly supported divergence free smooth function $\varphi$ and almost all $t > 0$, we may write

$$\langle u(t), \varphi \rangle - \langle u_0, \varphi \rangle = \int_0^t \langle g, \varphi\rangle_{V^* \times V} \tag{1}$$

where $g$ is a linear functional on $V$ defined by: $$g(\varphi) = -\langle \nabla u, \nabla \varphi \rangle - \langle(u \cdot \nabla )u, \varphi \rangle.$$

After a series of estimates (details are in this post: Trouble bounding a linear functional) , it is shown that $g \in L^{4/3}(0, T; V^*)$. The authors then claim that using the following result

Corollary 1.32 Suppose that $u, g \in L^1(0, T; X)$ and that for every $f \in X^*$ $$\langle f, u(t_2)\rangle - \langle f, u(t_1)\rangle = \int_{t_1}^{t_2} \langle f, g(s) \rangle ds$$ for almost every pair $(t_1, t_2)$ with $0 \leq t_1 \leq t_2 \leq T$. Then $\partial_t u = g$ in the sense of Definition 1.30 (which gives the definition for a weak time derivative, which I have omitted in this post).

we can conclude that $\partial_t u \in L^{4/3}(0,T; V^*)$.

I am trying to understand this result as I do not see how we justify using Corollary 1.32. It seems we are applying the corollary to (1) and viewing $\varphi$ as a linear functional on $V$? If so, I see how $g \in L^{4/3}(0, T; V^*) \implies g \in L^{1}(0, T; V^*)$ since $(0,T)$ is finite. I have two questions:

1. To use the corollary we would also require that $u \in L^{1}(0, T; V^*)$ but I do not see how this follows from anything done so far.
2. The corollary requires us to show that (1) holds for all $t_1$ and $t_2$, not just $t_1 = 0$ and $t_2 = t$ that appears in (1). How would one show this? My guess is this follows from Lemma 3.6.