A topology $\tau$ on $ L([0,1])$ that induces the $almost-everywhere$ convergence on $[0,1]$

Question

Let be $L([0,1])$ the vector space of equivalence classes of measurable function $f:[0,1]\rightarrow \mathbb{R}$ (I don't have more details about the type of equivalence that the exercise is talking abut, I guess it's this one: $[f]=\{g:[0,1]\rightarrow \mathbb{R}\ | g=f \ almost \ everywhere \ in \ [0,1]\}$). $$$$ Prove that: $\nexists \tau$ topology on $ L([0,1])$ that induces the $almost-everywhere$ convergence on $[0,1]$ $$$$ I don't have further details about what this means, I guess that if a sequence of equivalence class of functions $\{[f]_n\}\subset L([0,1])$ converge to a class of function $[f]\in L([0,1])$ then $f_n$ should converges to $f$ almost-everywhere on $[0,1]$. I wanted to prove this by absurd, creating a sequence of function that converges in $L([0,1])$ but not $almost-everywhre$ on $[0,1]$, any help?