# why does $f = \sum_{k}f_k$ exist?

The following theorem is taken from the book 'An Introductory Course in Functional Analysis' by Bowers and Kalton, page $25.$

Theorem $2.26:$ Let $(\Omega,\Sigma,\mu)$ be a positive measure space. The space $L_{\infty}(\Sigma,\mu)$ is a Banach space when given the essential supremum norm $\| \cdot \|_{\infty}.$

Proof: Again we use Lemme $2.24$ (Cauchy Summability Criteria). Let $(f_k)_{k}$ be a sequence of functions in $L_{\infty}(\Omega,\mu)$ and let $M = \sum_k\| f_k\|_{\infty}.$ Suppose $M<\infty.$ By assummption, for each $k \in \mathbb{N},$ we have $|f_k| \leq \| f_k \|_{\infty}$ a.e. $(\mu).$ Therefore, for each $k \in \mathbb{N},$ there exists a measurable set $N_k$ such that $\mu(N) = 0$ and $|f_k(\omega)| \leq \| f_k\|_{\infty}$ for all $\omega \in \Omega \setminus N_k.$ Let $N = \cup_{k}N_k$. Then $\mu(N) = 0$ and $f_k(\omega) < \infty$ for all $\omega \in \Omega \setminus N$ and all $k \in \mathbb{N}.$ Consequently, $f = \sum_{k}f_k$ exists a.e. and ....

Question: why does $f = \sum_{k}f_k$ exist? Do we need to show that $f$ is the limit of $\sum_{k=1}^nf_k$ in the essential supremum norm?

For first question, I think $f$ exists due to the Weierstrass M-test.

I ask the second question because the authors actually prove it. However, I thought we have proven that $f$ is the limit of $\sum_{k=1}^nf_k$ from the proof above?

• The function $f$ is the point wise limit $\sum_k f_k$ however using the Cauchy Summability Criteria you want $\sum_k f_k$ to converge to something using the norm of the space which is the essnetial supremum norm – clark Jun 8 '17 at 1:10

I agree that the authors worded it kind of confusingly. The authors prove that for every $\omega \in \Omega \setminus N$, the sum converges $\sum_{k=1}^\infty f(\omega)$ converges. Thus if you define $f(\omega) = \sum_{k=1}^\infty f_k(\omega)$, the function $f$ is a.e. everywhere defined. Up to this point the authors have only proven that $\sum_{k=1}^n f_k$ converges "pointwise" to $f$. So they still need to an argument to prove that it indeed converges in norm too.

Of course you could avoid this altogether by using the M-test. Since the M-test says this series will not only converge but also converge uniformly. And uniform convergence is equivalent to convergence in the supremum norm.