description/proof of that \(L^p\) over measure space is complete
Topics
About: metric space
The table of contents of this article
Starting Context
- The reader knows a definition of \(L^p\) over measure space.
- The reader knows a definition of metric induced by norm on real or complex vectors space.
- The reader knows a definition of complete metric space.
- The reader knows a definition of almost-everywhere over measure space.
- The reader knows a definition of extended Euclidean topological space.
- The reader knows a definition of Borel \(\sigma\)-algebra of topological space.
- The reader knows a definition of measure subspace of measure space for measurable subset.
- The reader admits the proposition that for any metric space, any Cauchy sequence into the metric space, and any positive number, there is a subsequence such that for each \(j\), the distance between the \(j\)-th element and the '\(j + 1\)'-th element is smaller than the number to the '\(j + 1\)'-th power.
- The reader admits the proposition that for any map between any measurable spaces, if the preimage of each element of any generator of the codomain \(\sigma\)-algebra is measurable, the map is measurable.
- The reader admits the monotone convergence theorem for Lebesgue integral.
- The reader admits the dominated convergence theorem for Lebesgue integral.
- The reader admits the proposition that for any Cauchy sequence on any metric space, if a subsequence of it converges to a point, the sequence converges to the same point.
- The reader admits the proposition that for any measurable map between any measurable spaces, the restriction on any domain subspace and any codomain subspace is measurable.
- The reader admits the proposition that for any map between any measurable spaces and any countable cover of the domain by any elements of the \(\sigma\)-algebra, if each domain restriction is measurable, the map is measurable.
- The reader admits the proposition that for any measurable space generated by any set of subsets and any subset, the subspace \(\sigma\)-algebra of the subset is generated by the set of the restricted subsets.
- The reader admits the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable.
- The reader admits the proposition that for any measure space, the union of any sequence of any locally negligible subsets is locally negligible.
- The reader admits the proposition that any uniformly Cauchy sequence of maps from any set into any complete metric space converges uniformly.
- The reader admits the proposition that for any measure space, any subset of any locally negligible subset of the space is locally negligible.
- The reader admits the proposition that for any uniformly convergent sequence of any value-bounded maps into any normed vectors space with the induced metric, the convergence is value-bounded.
- The reader admits the proposition that for any map from any measure space into the \(1\)-dimensional complex Euclidean topological space with the Borel \(\sigma\)-algebra, if and only if the map is measurable, the real and imaginary parts maps are measurable.
Target Context
- The reader will have a description and a proof of the proposition that \(L^p\) over any measure space is complete.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\((M, A, \mu)\): \(\in \{\text{ the measure spaces }\}\)
\(\mathbb{C}\): \(= \text{ the complex Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(F\): \(\in \{\mathbb{C}, \mathbb{R}\}\)
\(p\): \(\in \mathbb{R}\) such that \(1 \le p \lt \infty\)
\(L^p (M, A, \mu, F)\): with the metric induced by the norm
\(L^\infty (M, A, \mu, F)\): with the metric induced by the norm
//
Statements:
\(L^p (M, A, \mu, F) \in \{\text{ the complete metric spaces }\}\)
\(\land\)
\(L^\infty (M, A, \mu, F) \in \{\text{ the complete metric spaces }\}\)
//
2: Proof
Whole Strategy: Step 1: think of any Cauchy sequence, \(s: \mathbb{N} \to L^p (M, A, \mu, F)\), and one of its \(\mathcal{L}^p (M, A, \mu, F)\) versions, \(s': \mathbb{N} \to \mathcal{L}^p (M, A, \mu, F)\), and find a subsequence, \(s \circ f\), such that \(\Vert s \circ f (j + 1) - s \circ f (j) \Vert \lt (1 / 2)^{j + 1}\); Step 2: define \(g: M \to [0, \infty] := (lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert)^p\), and see that \(g\) is integrable; Step 3: see that \(lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) - s' \circ f (j)\) converges almost everywhere to an \(h\); Step 4: see that \(s \circ f\) converges to \([h]\); Step 5: think of any Cauchy sequence, \(s: \mathbb{N} \to L^\infty (M, A, \mu, F)\), and one of its \(\mathcal{L}^\infty (M, A, \mu, F)\) versions, \(s': \mathbb{N} \to \mathcal{L}^p (M, A, \mu, F)\), and define for each \(j \in \mathbb{N} \setminus \{0\}\), an \(N_j\) such that for each \(N_j \lt n, p\), \(\Vert s (n) - s (p) \Vert \lt 1 / j\), and each \(N_j \lt n, p\), \(S_{j, (n, p)} := \{m \in M \vert 1 / j \lt \vert (s' (n) - s' (p)) (m) \vert\}\) and define \(S_j := \cup_{N_j \lt n, p} S_{j, (n, p)}\) and \(S := \cup_j S_j\); Step 6: see that \(\{m \in M \vert 1 / j \lt \vert s' (n) (m) - h (m) \vert\}\) is locally negligible.
Step 1:
Let \(s: \mathbb{N} \to L^p (M, A, \mu, F)\) be any Cauchy sequence.
Each \(s (j)\) is an equivalence class of some elements of \(\mathcal{L}^p (M, A, \mu, F)\), and let us choose a representative element, \(s' (j) \in \mathcal{L}^p (M, A, \mu, F)\), \(s (j) = [s' (j)]\), so, we have a sequence, \(s': \mathbb{N} \to \mathcal{L}^p (M, A, \mu, F)\): use the axiom of choice.
There is an \(f: \mathbb{N} \to \mathbb{N}\) such that \(s \circ f: \mathbb{N} \to L^p (M, A, \mu, F)\) is a subsequence of \(s\) such that for each \(j \in \mathbb{N}\), \(\Vert s \circ f (j + 1) - s \circ f (j) \Vert \lt (1 / 2)^{j + 1}\), by the proposition that for any metric space, any Cauchy sequence into the metric space, and any positive number, there is a subsequence such that for each \(j\), the distance between the \(j\)-th element and the '\(j + 1\)'-th element is smaller than the number to the '\(j + 1\)'-th power.
\(lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \Vert s \circ f (j + 1) - s \circ f (j) \Vert \lt lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} (1 / 2)^{j + 1} = lim_{l \to \infty} ((1 / 2)^1 - (1 / 2)^{l + 2}) / (1 - 1 / 2) = ((1 / 2)^1) / (1 - 1 / 2) \lt \infty\).
Step 2:
Let us define \(g: M \to [0, \infty] := (lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert)^p\), where \([0, \infty]\) is the measurable subspace of \(\overline{\mathbb{R}}\) with the Borel \(\sigma\)-algebra (refer to the definition of extended Euclidean topological space).
Let us see that \(g\) is measurable.
\(s' \circ f (j + 1) - s' \circ f (j)\) is measurable.
\(\vert s' \circ f (j + 1) - s' \circ f (j) \vert\) is measurable, as a well known fact.
\(\sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert\) is measurable.
\(lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert\) converges over whole \(M\) (allowing \(\infty\) convergence) and is measurable as \(: M \to [0, \infty]\), as a well known fact.
\((lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert)^p\) is measurable, because \(f': [0, \infty] \to [0, \infty], r \mapsto r^p\) is measurable, because \([0, \infty]\) has the \(\sigma\)-algebra generated by the subspace topology of \(\overline{\mathbb{R}}\), by the proposition that for any measurable space generated by any set of subsets and any subset, the subspace \(\sigma\)-algebra of the subset is generated by the set of the restricted subsets, and each open subset of \([0, \infty]\), \(U' \subseteq [0, \infty]\), is \(U\) or \(U \cup \{\infty\}\) where \(U \subseteq [0, \infty)\) is an open subset, and \(f'^{-1} (U')\) is \({f' \vert_{[0, \infty)}}^{-1} (U)\) or \({f' \vert_{[0, \infty)}}^{-1} (U) \cup \{\infty\}\), which is open (because \(f' \vert_{[0, \infty)}\) is continuous), and by the proposition that for any map between any measurable spaces, if the preimage of each element of any generator of the codomain \(\sigma\)-algebra is measurable, the map is measurable, \(f'\) is measurable, and the proposition that for any measurable maps between any arbitrary subspaces of any measurable spaces, the composition is measurable applies.
Let us see that \(g\) is integrable.
\(\int_M g d \mu = \int_M (lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert)^p d \mu = \int_M lim_{l \to \infty} (\sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert)^p d \mu\), because the power function is continuous, \(= lim_{l \to \infty} \int_M (\sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert)^p d \mu\), by the monotone convergence theorem for Lebesgue integral, \(= lim_{l \to \infty} \Vert \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert \Vert^p\): \(s' \circ f (j + 1), s' \circ f (j) \in \mathcal{L}^p (M, A, \mu, F)\), \(s' \circ f (j + 1) - s' \circ f (j) \in \mathcal{L}^p (M, A, \mu, F)\) (it is a vectors space), \(\vert s' \circ f (j + 1) - s' \circ f (j) \vert \in \mathcal{L}^p (M, A, \mu, F)\) (a well-known fact), and \(\sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert \in \mathcal{L}^p (M, A, \mu, F)\) (it is a vectors space).
\(\le lim_{l \to \infty} (\sum_{j \in \{0, ..., l\}} \Vert \vert s' \circ f (j + 1) - s' \circ f (j) \vert \Vert)^p = lim_{l \to \infty} (\sum_{j \in \{0, ..., l\}} \Vert s' \circ f (j + 1) - s' \circ f (j) \Vert)^p = (lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \Vert s' \circ f (j + 1) - s' \circ f (j) \Vert)^p\), because the power function is continuous, \(\lt \infty\): \(\Vert s' \circ f (j + 1) - s' \circ f (j) \Vert = \Vert [s' \circ f (j + 1) - s' \circ f (j)] \Vert = \Vert [s' \circ f (j + 1)] - [s' \circ f (j)] \Vert = \Vert s \circ f (j + 1) - s \circ f (j) \Vert\) by definition.
Step 3:
Step 2 implies that \((lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert)^p \lt_{a.e.} \infty\), where let it be over \(M \setminus N\) where \(\mu (N) = 0\).
So, \(lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert \lt_{\text{ over } M \setminus N} \infty\), which implies that \(lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) - s' \circ f (j)\) converges in \(F\) over \(M \setminus N\), as is well known.
Let \(h: M \to F, m \mapsto s' \circ f (0) (m) + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) (m) - s' \circ f (j) (m) \text{ when } m \in M \setminus N; \mapsto 0 \text{ when } m \in N\).
Let us see that \(h\) is measurable.
We use the proposition that for any map from any measure space into the \(1\)-dimensional complex Euclidean topological space with the Borel \(\sigma\)-algebra, if and only if the map is measurable, the real and imaginary parts maps are measurable without further mentioning it.
\(h \vert_{M \setminus N}\) is measurable, because while \(s' \circ f (0) \vert_{M \setminus N}\) and \((\sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) - s' \circ f (j)) \vert_{M \setminus N}\) are measurable, by the proposition that for any measurable map between any measurable spaces, the restriction on any domain subspace and any codomain subspace is measurable, \((lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) - s' \circ f (j)) \vert_{M \setminus S}\) is measurable, as a well known fact.
\(h \vert_N\) is measurable, obviously.
So, by the proposition that for any map between any measurable spaces and any countable cover of the domain by any elements of the \(\sigma\)-algebra, if each domain restriction is measurable, the map is measurable, \(h\) is measurable.
\(\vert h \vert^p\) is integrable, because over \(M \setminus N\), \(\vert h \vert = \vert s' \circ f (0) (m) + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) (m) - s' \circ f (j) (m) \vert = \vert lim_{l \to \infty} (s' \circ f (0) (m) + \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) (m) - s' \circ f (j) (m)) \vert = lim_{l \to \infty} \vert s' \circ f (0) (m) + \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) (m) - s' \circ f (j) (m) \vert\), because the taking-absolute-value map is continuous, \(\le lim_{l \to \infty} (\vert s' \circ f (0) (m) \vert + \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) (m) - s' \circ f (j) (m) \vert) = \vert s' \circ f (0) (m) \vert + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) (m) - s' \circ f (j) (m) \vert\), and over \(N\), \(\vert h \vert = 0 \le \vert s' \circ f (0) (m) \vert + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) (m) - s' \circ f (j) (m) \vert\), but \(\vert s' \circ f (0) (m) \vert, lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) (m) - s' \circ f (j) (m) \vert \in \mathcal{L}^p (M, A, \mu, F)\), because \(g\) is integrable, and so, \(\vert s' \circ f (0) (m) \vert + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) (m) - s' \circ f (j) (m) \vert \in \mathcal{L}^p (M, A, \mu, F)\).
So, \(h \in \mathcal{L}^p (M, A, \mu, F)\).
Over \(M \setminus N\), \(h = s' \circ f (0) + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) - s' \circ f (j) = lim_{l \to \infty} (s' \circ f (0) + (s' \circ f (1) - s' \circ f (0)) + (s' \circ f (2) - s' \circ f (1)) + ... + (s' \circ f (l + 1) - s' \circ f (l))) = lim_{l \to \infty} s' \circ f (l + 1)\), so, over \(M \setminus N\), \(s' \circ f\) converges.
Step 4:
Let us see that \(s \circ f\) converges to \([h]\).
Over \(M \setminus N\), \(\vert h - s' \circ f (n) \vert = \vert s' \circ f (0) + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) - s' \circ f (j) - s' \circ f (n) \vert = \vert lim_{l \to \infty} (s' \circ f (0) + \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) - s' \circ f (j) - s' \circ f (n)) \vert = lim_{l \to \infty} \vert (s' \circ f (0) + \sum_{j \in \{0, ..., l\}} s' \circ f (j + 1) - s' \circ f (j) - s' \circ f (n)) \vert\), because the taking-absolute-value map is continuous, \( \le lim_{l \to \infty} (\vert s' \circ f (0) \vert + \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert + \vert s' \circ f (n) \vert) = \vert s' \circ f (0) \vert + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert + \vert s' \circ f (n) \vert\), but \(\vert s' \circ f (0) \vert, lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert, \vert s' \circ f (n) \vert \in \mathcal{L}^p (M, A, \mu, F)\), because \(lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert\) is measurable as is well known and \(g\) is integrable, so, \(\vert s' \circ f (0) \vert + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert + \vert s' \circ f (n) \vert \in \mathcal{L}^p (M, A, \mu, F)\).
So, \(\vert h - s' \circ f (n) \vert^p \le (\vert s' \circ f (0) \vert + lim_{l \to \infty} \sum_{j \in \{0, ..., l\}} \vert s' \circ f (j + 1) - s' \circ f (j) \vert + \vert s' \circ f (n) \vert)^p\), where the right hand side is integrable over \(M\), so over \(M \setminus N\).
So, by the dominated convergence theorem for Lebesgue integral, \(lim_{n \to \infty} \int_{M \setminus N} \vert h - s' \circ f (n) \vert^p d \mu = \int_{M \setminus N} lim_{n \to \infty} \vert h - s' \circ f (n) \vert^p d \mu\).
But over \(M \setminus N\), \(h = lim_{l \to \infty} s' \circ f (l + 1)\), so, over there, \(lim_{n \to \infty} \vert h - s' \circ f (n) \vert^p = lim_{n \to \infty} \vert lim_{l \to \infty} s' \circ f (l + 1) - s' \circ f (n) \vert^p = 0\).
So, \(\int_M lim_{n \to \infty} \vert h - s' \circ f (n) \vert^p d \mu = \int_{M \setminus N} lim_{n \to \infty} \vert h - s' \circ f (n) \vert^p d \mu + \int_{N} lim_{n \to \infty} \vert h - s' \circ f (n) \vert^p d \mu = \int_{M \setminus N} 0 \mu + \int_{N} lim_{n \to \infty} \vert h - s' \circ f (n) \vert^p d \mu = 0\), because \(\mu (N) = 0\).
So, \(lim_{n \to \infty} \int_M \vert h - s' \circ f (n) \vert^p d \mu = 0\).
That means that \(lim_{n \to \infty} \Vert h - s' \circ f (n) \Vert = 0\), which means that \(lim_{n \to \infty} \Vert [h - s' \circ f (n)] \Vert = lim_{n \to \infty} \Vert [h] - [s' \circ f (n)] \Vert = lim_{n \to \infty} \Vert [h] - s \circ f (n) \Vert = 0\), which means that \(s \circ f\) converges to \([h]\).
By the proposition that for any Cauchy sequence on any metric space, if a subsequence of it converges to a point, the sequence converges to the same point, \(s\) converges to \([h]\).
Step 5:
Let \(s: \mathbb{N} \to L^\infty (M, A, \mu, F)\) be any Cauchy sequence.
Each \(s (j)\) is an equivalence class of some elements of \(\mathcal{L}^\infty (M, A, \mu, F)\), and let us choose a representative element, \(s' (j) \in \mathcal{L}^\infty (M, A, \mu, F)\), \(s (j) = [s' (j)]\), so, we have a sequence, \(s': \mathbb{N} \to \mathcal{L}^\infty (M, A, \mu, F)\): use the axiom of choice.
For each \(j \in \mathbb{N} \setminus \{0\}\), there is an \(N_j \in \mathbb{N}\) such that for each \(n, p \in \mathbb{N}\) such that \(N \lt n, p\), \(\Vert s (n) - s (p) \Vert \lt 1 / j\).
\(\Vert s (n) - s (p) \Vert = \Vert s' (n) - s' (p) \Vert = inf \{r \in \mathbb{R} \vert 0 \le r \land \{m \in M \vert r \lt \vert (s' (n) - s' (p)) (m) \vert\} \in \{\text{ the locally negligible subsets of } M\}\}\).
Let \(S_{j, (n, p)} := \{m \in M \vert 1 / j \lt \vert (s' (n) - s' (p)) (m) \vert\}\), which is locally negligible, because \(\Vert s (n) - s (p) \Vert \lt 1 / j\).
\(S_{j, (n, p)} \in A\), because \(\vert (s' (n) - s' (p)) \vert\) is measurable and \(S_{j, (n, p)} = \vert (s' (n) - s' (p)) \vert^{-1} ((1 / j, \infty))\).
For each \(m \in M \setminus S_{j, (n, p)}\), \(\vert (s' (n) - s' (p)) (m) \vert \le 1 / j\).
Let \(S_j := \cup_{N_j \lt n, p} S_{j, (n, p)}\), which is locally negligible, by the proposition that for any measure space, the union of any sequence of any locally negligible subsets is locally negligible, because \(\{(n, p) \in \mathbb{N} \times \mathbb{N} \vert N_j \lt n, p\}\) is countable.
\(S_j \in A\).
Let \(S := \cup_j S_j\), which is locally negligible, by the proposition that for any measure space, the union of any sequence of any locally negligible subsets is locally negligible.
\(S \in A\).
Step 6:
Let us see that over \(M \setminus S\), \(s'\) is uniformly Cauchy.
Let \(\epsilon \in \mathbb{R}\) be any such that \(0 \lt \epsilon\).
There is a \(j \in \mathbb{N} \setminus \{0\}\) such that \(1 / j \lt \epsilon\).
For each \(m \in M \setminus S\), \(m \notin S_{j, (n, p)}\) for each \(j \in \mathbb{N} \setminus \{0\}\) and each \(n, p \in \mathbb{N}\) such that \(N_j \lt n, p\), so, \(m \in M \setminus S_{j, (n, p)}\).
So, \(\vert (s' (n) - s' (p)) (m) \vert \le 1 / j \lt \epsilon\) for each \(n, p \in \mathbb{N}\) such that \(N_j \lt n, p\), where \(N_j\) is independent of \(m\).
So, over \(M \setminus S\), \(s'\) is uniformly Cauchy.
So, \(s'\) over \(M \setminus S\) converges to an \(h: M \setminus S \to F\) uniformly, by the proposition that any uniformly Cauchy sequence of maps from any set into any complete metric space converges uniformly.
\(h: M \setminus S \to F\) is measurable as a well known fact: the limit of any sequence of measurable functions is measurable.
Let us expand that \(h\) to \(h: M \to F\) such that \(h (m) = 0\) for each \(m \in S\).
\(h \vert_S\) is obviously measurable.
\(h\) is measurable, because \(M \setminus S, S \in A\) and \(h \vert_{M \setminus S}\) and \(h \vert_S\) are measurable and the proposition that for any map between any measurable spaces and any countable cover of the domain by any elements of the \(\sigma\)-algebra, if each domain restriction is measurable, the map is measurable applies.
\(h\) is value-bounded, by the proposition that for any uniformly convergent sequence of any value-bounded maps into any normed vectors space with the induced metric, the convergence is value-bounded.
So, \(h \in \mathcal{L}^\infty (M, A, \mu, F)\).
For each \(j \in \mathbb{N} \setminus \{0\}\), there is an \(N \in \mathbb{N}\) such that for each \(n \in \mathbb{N}\) such that \(N \lt n\), for each \(m \in M \setminus S\), \(\vert s' (n) (m) - h (m) \vert \lt 1 / j\).
So, \(\{m \in M \vert 1 / j \lt \vert s' (n) (m) - h (m) \vert\} \subseteq S\), which is locally negligible, by the proposition that for any measure space, any subset of any locally negligible subset of the space is locally negligible, because \(S\) is locally negligible.
That means that \(\Vert s' (n) - h \Vert \le 1 / j\).
That means that \(\Vert s' (n) - h \Vert\) converges to \(0\).
So, \(\Vert s' (n) - h \Vert = \Vert [s' (n) - h] \Vert = \Vert [s' (n)] - [h] \Vert = \Vert s (n) - [h] \Vert\) converges to \(0\).
That means that \(s\) converges to \([h]\).
So, \(L^\infty (M, A, \mu, F)\) is complete.
