$\ell^p(J)$ is complete (Banach)

I have to prove that the space

$$\ell^p(J)$$ defined as the set of all functions

$$\psi: J\rightarrow \mathbb{F}$$

s.t. $$\psi$$ is null except in a contable subset of $$J$$

and

$$||\psi||_p :=\bigg(\sum_{t\in J}|\psi(t)|^p\bigg)^{1/p}<\infty$$

is Banach (complete).

Well, given a Cauchy sequence $$(\psi_n)_n\subset \ell^p(J)$$ I defined $$\psi$$ such that each $$\psi(t)$$ is defined as $$\lim \psi_n(t)$$ (each of these sequences $$\psi_n(t)$$ is Cauchy in $$\mathbb{F}$$).

I proved that $$\psi_n\rightarrow \psi$$ and $$||\psi||_p<\infty$$. However, I could not prove that $$\psi$$ is null except in a contable subset of $$J$$.

Many thanks for any help.

Hint: if $$\psi(t) \ne 0$$, then some $$\psi_n(t) \ne 0$$.
• Yes, for $n$ bigg... But combining various $n$'s biggs, I could not prove yet the result. Many thanks for the help. – Na'omi Apr 23 at 16:38
• $\psi (t) \neq 0 \implies \psi_n (t) \neq 0$ to $n$ big enough. Thus, where $J_n$ is the subset of $J$ with zero measure out of which $\psi_n$ cancels, the points on which $\psi$ does not vanish are contained in the enumerable union $\cup J_n$, also measure null... Is it correct? Many thanks. – Na'omi Apr 23 at 18:10