I want to show that the sequence $a_n(t)=\frac{1}{n}\textbf{1}_{[0,n)}(t)$ does not converge in $L^1(\lambda)$. $u_n$ converges in $L^p$ implies $u_n$ is Cauchy so I want to use the contrapositive.
$$\begin{align}||a_{2n}-a_n||_1&=\int|a_{2n}-a_n|~d(\lambda)\\&=\int\left|\frac{1}{2n}\textbf{1}_{[0,2n)}(t)-\frac{1}{n}\textbf{1}_{[0,n)}(t)\right|~d(\lambda)\\&=\int\left|\frac{1}{n}\left(\frac{1}{2}\textbf{1}_{[0,2n)}(t)-\textbf{1}_{[0,n)}(t)\right)\right|~d(\lambda) \end{align} $$
not sure how to continue
I know the sequence does not converge. I want to prove it doesn't; this is why the question isn't a duplicate.
You are on the right track. Notice that $$\int_0^{+\infty}\left|\frac{1}{n}\left(\frac{1}{2}\textbf{1}_{[0,2n)}(t)-\textbf{1}_{[0,n)}(t)\right)\right|~d(\lambda)= \frac{1}{n}\int_0^{n}\left|-\frac{1}{2}\right|~d(\lambda) +\frac{1}{n}\int_n^{2n}\left|\frac{1}{2}\right|~d(\lambda)=1$$ which does not go to zero as $n\to+\infty$.