$u_n \in L^1$ $u_n \to u$ in $L^1$ and $\int_0^1 u_n \ dt = 1$ $\implies$ $\int_0^1 u \ dt = 1$ Consider a sequence $u_n \in L^1([0,1])$. Suppose that $u_n \to u$ in $L^1([0,1])$.
If $$\int_0^1 u_n \ dt = 1$$ $\forall n \in \mathbb{N}$ then why $$\int_0^1 u \ dt = 1?$$ I'm sure this follows from a very basic result, but I'm not able to figure it out right now.

Note that $u_n \to u$ means that $(u_n - u) \to 0$ in $L^1$, which is to say that $\int|u_n - u| \to 0$.  We want to show that $\int u_n - \int u \to 0$. Note, however, that
$$
\left| \int u_n - \int u\right| =
\left| \int (u_n - u)\right| \leq \int |u_n - u|
$$
