Let $\{\mu_t\}_{t\in [0, T]}$ measures on the interval $[0,1]$. Suppose that
$$\sup_{\phi\in L^1[0,1], \|\phi\|\leq 1, \phi\geq 0}\int_0^Tdt\int_0^1\phi(x)d\mu_t(x)<+\infty$$.
Is this condition enough to conclude that the measure $\mu_t$ is absolutely continuous respect to the Lesbegue measure for a.e. $t\in [0, T]$? Reading some notes I found a step like that but I don't understand why and I am not even sure that I understood correctly. Could someone help me?
As pointed out in the comments, it is possible that the integral of $\phi$ against $\mu_t$ may not even be well defined if $\mu_t$ isn't absolutely continuous with respect to the Lebesgue measure. This is because $\phi \in L^1[0,1]$ is an equivalence class of functions which are equal to each other Lebesgue almost everywhere. This condition is not enough to imply that the integrals against $\mu_t$ of every member of that equivalence class is the same.
One strong assumption you could make to get around this is to assume that the $\sup$ is finite even when taken not just over non-negative elements of $L^1[0,1]$ but instead over all functions that lie in the equivalence class of a non-negative element of $L^1[0,1]$.
Even this condition is not enough to guarantee absolute continuity of $\mu_t$ with respect to the Lebesgue measure for a single $t$ in $[0,1]$, nevermind a.e. $t$. For simplicity I fix $T = 1$ here.
For each $t$ let $\mu_t = \delta_t$ be the dirac measure at $t$. Then for $\phi$ in the equivalence class of some non-negative element of $L^1[0,1]$, $$\int_0^1 \phi(x) d\mu_t(x) = \phi(t).$$
Hence we have that
$$\int_0^1 \int_0^1 \phi(x) d\mu_t(x) dt = \|\phi\|_{L^1[0,1]}$$ for all $\phi \in L^1[0,1]$ such that $\phi \geq 0$ so that your $\sup$ is finite. However, it is clear that there is no $t$ such that $\mu_t$ is absolutely continuous with respect to the lebesgue measure.