Weak continuity in time for a solution to the continuity equation

Let $$\Omega = (0,1)$$. Suppose $$\rho \in L^{\infty}(0,T; L^{\infty}(\Omega))$$ satisfies the following weak formulation of the continuity equation

$$\int_{0}^{T} \int_{\Omega} \rho(t,x) \partial_{t}\varphi + \rho(t,x)u(t,x) \partial_{x}\varphi(t,x)~dxdt = 0,$$ for all $$\varphi \in C^{\infty}_{c}((0,T) \times \Omega)$$. Also assume $$\rho u \in L^{2}(0,T; L^{2}(\Omega))$$.

Apparently in the book I am reading it follows that the function $$t \mapsto \int_{\Omega} \rho(t,x)\varphi(x)~dx$$ is absolutely continuous in $$[0,T]$$. But I don't see why.

My attempt is to show that $$\partial_{t} \int_{\Omega} \rho \varphi~dx \in L^{1}(0,T)$$. But this would require us to integrate by parts in the first term of the weak formulation. However $$\rho$$ is not differentiable so we cant do that in the classical sense. But if we understand the derivative in the weak sense then we could say

$$\int_{\Omega} \partial_{t} \rho \cdot\varphi~dx = \partial_{t} \int_{\Omega} \rho \varphi~dx \in L^{1}(0,T)$$ using the weak formulation and integrability of $$\rho u$$. But then I am unsure if this swapping of the derivative and integral can be justified rigorously or not.

Define, for fixed $$\varphi_2\in C_c^\infty(\Omega)$$, $$F(t):=\int_{\Omega} \rho(t,x)\varphi_2(x)\, dx$$ We want to show that $$F$$ is absolutely continuous on $$[0,T]$$; we will in fact show that $$F\in W^{1,2}(0,T)$$.
First we note that $$F$$ is bounded: For a.e. $$t\in (0,T)$$, $$|F(t)|\leq \Vert \rho\Vert_{L^\infty((0,T);L^\infty( \Omega))} \Vert \varphi_2\Vert_{L^1(\Omega)}<\infty.$$
Next, to show that $$F$$ has a weak derivative in $$L^2(0,T)$$, it's enough to show (this is in Brezis' functional analysis book) that there exists a constant $$C>0$$ such that for every $$\varphi_1\in C_c^\infty(0,T)$$ we have $$\left|\int_0^T F(t)\varphi_1'(t)\, dt \right| \leq C\| \varphi_1\|_{L^2(0,T)}.$$ However, note that if we define $$\varphi(t,x):=\varphi_1(t)\varphi_2(x)$$, then $$\varphi$$ is an admissible test function in the weak formulation and $$\int_0^T F(t)\varphi_1'(t)\, dt = \int_0^T\int_\Omega \rho(t,x)\partial_t \varphi(t,x)\, dx dt = -\int_0^T\int_\Omega \rho(t,x)u(t,x)\partial_x\varphi_2(x)\varphi_1(t)\, dxdt.$$ Using the Cauchy-Schwarz inequality, we see (remember that $$\rho, u, \varphi_2$$ are fixed!) $$\left|\int_0^T F(t)\varphi_1'(t)\, dt \right|\leq \| \rho u\|_{L^2((0,T);L^2(\Omega))} \| \partial_x\varphi_2\|_{L^2(\Omega)}\| \varphi_1\|_{L^2(0,T)}=:C\| \varphi_1\|_{L^2(0,T)}.$$