Given norm $\| \cdot \|'$ on $L^1(\mathbb{R})$ such that $L^1(\mathbb{R})$ is still banach and from $f_n \to f$ in $\| \cdot \|'$ follows that $\int_{-\infty}^t f_n(s)ds \to \int_{-\infty}^t f(s)ds$ for any $t$. I need to show that this norm is equivalent to usual norm on $L^1(\mathbb{R})$.
My idea is that first of all $\phi_t(x) = \int_{-\infty}^t x(s)ds$ is a linear functional and it is continuous with respect to $\| \cdot \|'$ because it preserves convergence of sequences. Next I claim (and it is not correct but I hope it is close to correct) that $\sup_t \phi_t(x) = \|x\|$ (It's $L^1$-norm and that's where I'm lying) $ = \sup_t \|\phi_t(x)\|\|x\|' = C\|x\|'$ by the Uniform Boundedness Principle. And that's pretty much it except that I don't know how to make $\sup_t \phi_t(x) = \|x\|$ to be true because actually I should have had $\sup_t \phi_t(|x|) = \|x\|$ but it would have spoiled everything else.