Let $u = u(t,x)$, where $(t,x) \in [0,T] \times \mathbb{R}^d$ and for a fixed $1 \le p < \infty$, $ u(t,\cdot) \in L^p(\mathbb{R}^d)$ for all $t \in [0,T]$.
My question is,
if $\Vert u(t,\cdot)\Vert_{L^p} = 0 $ a.e. $ t\in [0,T]$, then $u \equiv 0$ on $[0,T]\times \mathbb{R}^d$ a.e.? Please share your wisdom. Thanks!
Assuming $(t,x)\to u(t,x)$ is measurable (this can be slightly relaxed) then yes.
This is exactly what Fubini-Tonelli is for.
Let $A= \{(t,x): u(t,x)=0\}$. Then
$$(\lambda\times \lambda) (A) = \iint {\bf 1}_A (t,x) d \lambda \times \lambda = \int_{[0,T]} \left(\int_{{\mathbb R}^d} {\bf 1}_A (t,x) d\lambda (x)\right) d \lambda (t)=0.$$
The first equality is by definition, the second by Fubini-Tonelli (which also claims that the inner integral is a measurable function). By assumption the inner integral is zero a.e., and therefore the entire integral is zero, giving the last equality.