Let $X_n, n \geq 1$ be random variables. Suppose that $X_n \rightarrow 0$ $\mathbb P$-a.s. as $n \rightarrow \infty$. Moreover, suppose that $E[X_n]=1$.
Now is it clear that there cannot be $L^1$-convergence? Why? Because the limit would need to be 0, whereas the mean converges to 1? But convergence of the mean hasn't directly to do with $L^1$-convergence, because for that we would consider the mean of the difference?
So why can we conclude that we do not have $L^1$-convergence? What would we need to have $L^1$-convergence?
As David hints, $L^1$ convergence implies almost sure convergence of a subsequence to the same limit. So if $X_n$ converges in $L^1$, the limit must be zero as you stated. Then noting that $E|X_n| \to 0$ (by $L^1$ convergence) and $E|X_n| \ge E X_n=1$, we have a contradiction.