I have shown that if $(V,||\cdot||)$ is a normed space in which every absolutely convergent series converges, then it is in fact a Banach space.
I now need to apply this lemma to the Lebesgue space $L^1(X)$, defined to be the space of equivalence classes of $\hat{L^1}(X)$, where $\hat{L^1}(X)$ is the space of integrable functions on $X \subset \mathbb{R}$ and is equipped with the semi-norm $||\cdot||_{L^1}$, and two functions are equivalent if and only if they are equal almost everywhere. I have shown that every absolutely convergent series in $\hat{L^1}(X)$ is convergent, but when it comes to using equivalence classes and representatives I get muddled:
Suppose $\sum_{k=1}^\infty ||\zeta_k|| < \infty$, where $\zeta_k \in L^1(X)$. For each $k$, take some representative $f_k \in \hat{L^1}(X)$ of the equivalence class $\zeta_k$, so $\zeta_k = [f_k]$. Then, by the definition of the $L^1$ norm on the set of equivalence classes, we have $||\zeta_k|| = ||f_k||_{L^1}$ for each $n$, and so $\sum_{k=1}^\infty ||f_k||_{L^1} < \infty$ and therefore there exists an $f \in \hat{L^1}(X)$ such that $\sum_{k=1}^n f_k \rightarrow f$ in the $L^1$ norm. I now want to show that $\sum_{k=1}^n \zeta_k$ converges to the equivalence class $[f]$.
So given $\varepsilon>0$ there exists an $N$ such that $||\sum_{k=1}^n f_k – f||_{L^1} < \varepsilon$ for all $n>N$. Then I get stuck when I try and consider $||\sum_{k=1}^n \zeta_k – [f]||$, as we can only apply this norm to an equivalence class $[\cdot]$. I tried using the triangle inequality and bringing the $[f]$ into the sum by splitting it into $n$ separate $[\frac{1}{n}f]$ so that I could apply the norm but it didn't get me anywhere.
Any tips please?
Best Answer
$\sum\limits_{k=1}^{n} \zeta_k$ is nothing but the equivalence class of $\sum\limits_{k=1}^{n} f_k$ and $\|\sum\limits_{k=1}^{n} \zeta_k-[f]\|$ is nothing but $\|\sum\limits_{k=1}^{n} f_k-f\|_{L^{1}}$ which is less than $\epsilon$.