I was looking into the Navier-Stokes Weak formulation
Let $f\in L^2(\Omega_T)$, $u_0 \in H(\Omega)=\lbrace u\in L^2(\Omega):\text{div }u=0\text{ in }\Omega;u\cdot n|_{\partial \Omega}=0\rbrace$.
A measurable function $u:\Omega_T\rightarrow \mathbb{R}^N$, $N=2,3,$ is said to be a weak solution to Navier-Stokes equations in $\Omega_T$ if
$u \in V_T \equiv L^2(0,T;H^1_0)\cap L^{\infty}(0,T;H)$ and $u$ satisfies $$\int_0^{\infty}\Big(\big(u,\frac{\partial\varphi}{\partial t}\big)-\nu(\nabla u,\nabla\varphi)-(u\cdot\nabla u,\varphi)\Big)\text{ d}t=-\int_0^{\infty}(f,\varphi)\text{ d}t-(u_0,\varphi(0))$$
$$\forall \varphi\in \mathcal{D}_T=\lbrace \varphi \in C_0^{T}(\Omega_T):\; \text{div}\varphi(x,t)=0 \text{ in } \Omega_T\rbrace.$$
Then I've found this lemma, said to be useful
Let $u$ be a weak solution in $\Omega_T$. Then $u$ can be redefined on a set of zero Lebesgue measure in such a way that $u(t) \in L^2(\Omega)$ $\forall t\in[0,T)$ and satisfies \begin{align*} \int_s^t\Big(\big(u,\frac{\partial\varphi}{\partial t}\big)-\nu(\nabla u,\nabla\varphi)-(u\cdot\nabla u,\varphi)\Big)\text{ d}\tau\\ =-\int_s^{t}(f,\varphi)\text{ d}\tau+(u(t),\varphi(t))-(v(s),\varphi(s)),\\\forall s\in[0,t],t<T,\text{ and }\forall \varphi\in \mathcal{D}_T. \end{align*}
Then there are some theorems about equivalency of this formulations. But I don't really think I can understand well the main idea, so obviously we use smaller integration interval but how does this contribute to futher results.
I would be really grateful for some explanation, thanks!