Finite Measure Space: Integral Closure = Bochner Integral

Question

I can't sleep for so long time as the integral gives me headaches.
I was looking for so many approaches. Now another one. Hope this works...

Let $\Omega$ be a finite measure space and $E$ a Banach space.
Consider the function space $\mathcal{F}(\Omega,E)$ with the topology of $\mu$-almost everywhere convergence.
Define the integral of a a simple function $s=\sum_i a_i \chi_{A_i}$ as $I(s):=\sum_i a_i\mu(A_i)$.

Is the operator closable, i.e.: $$(s_n,I(s_n))\to(0,c)\implies(c=0)$$ (Note that the measure space is finite $\mu(\Omega)<\infty$, so there can't be running blocks to infinity...)

Moreover does the closure coincide with the Bochner integral?

Answer 1

Ok to bad, the answer is again: No!

For $s_n:=n\chi_{[0,\frac{1}{n}]}$ it follows $s_n\to 0$ but $I(s_n)\to 1$.