description/proof of that for measurable extended real function over measure space whose range is in union of natural numbers set and infinity, Lebesgue integral of map is sum of measures of preimages of closed-positive-natural-number-or-infinity-lower-bounded intervals
Topics
About: measure space
The table of contents of this article
Starting Context
- The reader knows a definition of Lebesgue integral of measurable extended real function over measurable subset of measure space.
- The reader admits the proposition that for any map, the map preimage of the range is the whole domain.
- The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
- The reader admits the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
- The reader admits the proposition that for any absolutely convergent double series on the \(1\)-dimensional Euclidean metric space, the series with the sums orders changed converge to the same convergence.
- The reader admits the proposition that for any divergent non-negative double series on the \(1\)-dimensional Euclidean metric space, the series with the sums orders changed diverges.
- The reader admits the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger.
Target Context
- The reader will have a description and a proof of the proposition that for any measurable extended real function over any measure space whose range is in the union of the natural numbers set and the infinity, the Lebesgue integral of the map is the sum of the measures of the preimages of the closed-positive-natural-number-or-infinity-lower-bounded intervals.
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:
\((M, A, \mu)\): \(\in \{\text{ the measure spaces }\}\)
\(f\): \(: M \to [0, \infty]\), \(\in \{\text{ the measurable maps }\}\), such that \(f (M) \subseteq \mathbb{N} \cup \{\infty\}\)
//
Statements:
\(\int_M f d \mu = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} n \mu (f^{- 1} (\{n\})) = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} \mu (f^{-1} ([n, \infty]))\)
//
2: Proof
Whole Strategy: Step 1: see that \(\int_M f d \mu = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} n \mu (f^{- 1} (\{n\}))\); Step 2: see that \(\sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} \mu (f^{-1} ([n, \infty])) = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} n \mu (f^{- 1} (\{n\}))\); Step 3: conclude the proposition.
Step 1:
Let \(h: M \to \mathbb{R}\) be any non-negative simple measurable function such that \(h \le f\).
Let \(h (M) = \{r_1, ..., r_l\}\).
Let us see that \(\int_M h d \mu = \sum_{j \in \{1, ..., l\}} r_j \mu (h^{-1} (\{r_j\})) \le \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (f^{- 1} (\{n\}))\).
\(h^{-1} (\{r_j\}) = h^{-1} (\{r_j\}) \cap M = h^{-1} (\{r_j\}) \cap f^{-1} (\mathbb{N} \cup \{\infty\})\), by the proposition that for any map, the map preimage of the range is the whole domain, \(= h^{-1} (\{r_j\}) \cap (\cup_{n \in \mathbb{N} \cup \{\infty\}} f^{-1} (\{n\}))\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets, \(= \cup_{n \in \mathbb{N} \cup \{\infty\}} (h^{-1} (\{r_j\}) \cap f^{-1} (\{n\}))\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset.
\(r_j \mu (h^{-1} (\{r_j\})) = r_j \mu (\cup_{n \in \mathbb{N} \cup \{\infty\}} (h^{-1} (\{r_j\}) \cap f^{-1} (\{n\}))) = r_j \sum_{n \in \mathbb{N} \cup \{\infty\}} \mu (h^{-1} (\{r_j\}) \cap f^{-1} (\{n\})) = \sum_{n \in \mathbb{N} \cup \{\infty\}} r_j \mu (h^{-1} (\{r_j\}) \cap f^{-1} (\{n\})) \le \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (h^{-1} (\{r_j\}) \cap f^{-1} (\{n\}))\), because for each \(n\) such that \(n \lt r_j\), \(h^{-1} (\{r_j\}) \cap f^{-1} (\{n\}) = \emptyset\), because for each \(m \in h^{-1} (\{r_j\})\), \(h (m) = r_j \le f (m) \neq n\), so, \(m \notin f^{-1} (\{n\})\), so, only the \(r_j \le n\) terms are nonzero.
So, \(\sum_{j \in \{1, ..., l\}} r_j \mu (h^{-1} (\{r_j\})) \le \sum_{j \in \{1, ..., l\}} \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (h^{-1} (\{r_j\}) \cap f^{-1} (\{n\})) = \sum_{n \in \mathbb{N} \cup \{\infty\}} \sum_{j \in \{1, ..., l\}} n \mu (h^{-1} (\{r_j\}) \cap f^{-1} (\{n\}))\), by the proposition that for any absolutely convergent double series on the \(1\)-dimensional Euclidean metric space, the series with the sums orders changed converge to the same convergence and the proposition that for any divergent non-negative double series on the \(1\)-dimensional Euclidean metric space, the series with the sums orders changed diverges, \(= \sum_{n \in \mathbb{N} \cup \{\infty\}} n \sum_{j \in \{1, ..., l\}} \mu (h^{-1} (\{r_j\}) \cap f^{-1} (\{n\})) = \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (\cup_{j \in \{1, ..., l\}} (h^{-1} (\{r_j\}) \cap f^{-1} (\{n\}))) = \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu ((\cup_{j \in \{1, ..., l\}} h^{-1} (\{r_j\})) \cap f^{-1} (\{n\})))\), by the proposition that for any set, the intersection of the union of any possibly uncountable number of subsets and any subset is the union of the intersections of each of the subsets and the latter subset, \(= \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (h^{-1} (\cup_{j \in \{1, ..., l\}} \{r_j\}) \cap f^{-1} (\{n\})))\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets, \(= \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (M \cap f^{-1} (\{n\}))\), by the proposition that for any map, the map preimage of the range is the whole domain, \(= \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (f^{-1} (\{n\}))\).
So, \(\sum_{j \in \{1, ..., l\}} r_j \mu (h^{-1} (\{r_j\})) \le \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (f^{- 1} (\{n\}))\).
If \(0 \lt \mu (f^{- 1} (\{\infty\}))\), \(\sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (f^{- 1} (\{n\})) = \infty\) while \(int_M f d \mu = \infty\), so, \(int_M f d \mu = \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (f^{- 1} (\{n\}))\).
Let us suppose otherwise hereafter.
\(\sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (f^{- 1} (\{n\})) = \sum_{n \in \mathbb{N}} n \mu (f^{- 1} (\{n\}))\).
Let us suppose that \(\sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (f^{- 1} (\{n\})) = \infty\).
For each large \(r \in \mathbb{R}\) such that \(0 \le r\), there is an \(N \in \mathbb{N}\) such that \(r \lt \sum_{n \in \{0, ..., N + 1\}} n \mu (f^{- 1} (\{n\}))\).
Then, there is the non-negative simple measurable function, \(f_{N + 1}: M \to \mathbb{R}, m \mapsto Min (\{f (m), N + 1\})\), such that \(f_{N + 1} \le f\).
\(f_{N + 1}\) is indeed non-negative, because \(0 \le f (m)\) and \(0 \le N + 1\).
\(f_{N + 1}\) is indeed simple, because \(f_{N + 1} (M) \subseteq \{0, ..., N + 1\}\).
Let us see that \(f_{N + 1}\) is indeed measurable.
Let \(a \subseteq \mathbb{R}\) be any measurable subset.
When \(N + 1 \notin a\), \({f_{N + 1}}^{-1} (a) = f^{-1} (a \cap (- \infty, N + 1))\), because for each \(m \in {f_{N + 1}}^{-1} (a)\), \(f_{N + 1} (m) \in a\), but \(f_{N + 1} (m) \in (- \infty, N + 1)\), so, \(f_{N + 1} (m) = Min (\{f (m), N + 1\}) = f (m) \in a \cap (- \infty, N + 1)\), so, \(m \in f^{-1} (a \cap (- \infty, N + 1))\); for each \(m \in f^{-1} (a \cap (- \infty, N + 1))\), \(f (m) \in a \cap (- \infty, N + 1)\), but \(f_{N + 1} (m) = Min (\{f (m), N + 1\}) = f (m) \in a \cap (- \infty, N + 1) \subseteq a\), so, \(m \in {f_{N + 1}}^{-1} (a)\).
When \(N + 1 \in a\), \({f_{N + 1}}^{-1} (a) = f^{-1} (a) \cup f^{-1} ([N + 1, \infty])\), because for each \(m \in {f_{N + 1}}^{-1} (a)\), \(f_{N + 1} (m) \in a\), but when \(f_{N + 1} (m) \lt N + 1\), \(f_{N + 1} (m) = Min (\{f (m), N + 1\}) = f (m) \in a\), so, \(m \in f^{-1} (a)\), and when \(f_{N + 1} (m) = N + 1\), \(f_{N + 1} (m) = Min (\{f (m), N + 1\}) = N + 1\), so, \(N + 1 \le f (m)\), so, \(m \in f^{-1} ([N + 1, \infty])\); for each \(m \in f^{-1} (a) \cup f^{-1} ([N + 1, \infty])\), when \(m \in f^{-1} (a)\), \(f (m) \in a\), and \(f_{N + 1} (m) = Min (\{f (m), N + 1\}) = f (m) \text{ or } N + 1\), and anyway, \(f_{N + 1} (m) \in a\), so, \(m \in {f_{N + 1}}^{-1} (a)\), and when \(m \in f^{-1} ([N + 1, \infty])\), \(f (m) \in [N + 1, \infty]\), but \(f_{N + 1} (m) = Min (\{f (m), N + 1\}) = N + 1 \in a\), so, \(m \in {f_{N + 1}}^{-1} (a)\).
So, anyway, \({f_{N + 1}}^{-1} (a)\) is measurable.
So, \(f_{N + 1}\) is measurable.
Then, \(\sum_{n \in \{0, ..., N + 1\}} n \mu (f^{- 1} (\{n\})) \le \sum_{n \in \{0, ..., N + 1\}} n \mu ({f_{N + 1}}^{- 1} (\{n\}))\), because \(f^{- 1} (\{n\}) \subseteq {f_{N + 1}}^{- 1} (\{n\})\), because for each \(m \in f^{- 1} (\{n\})\), \(f (m) = n \le N + 1\), so, \(f_{N + 1} (m) = Min (\{f (m), N + 1\}) = f (m) = n\), so, \(m \in {f_{N + 1}}^{- 1} (\{n\})\).
So, \(r \lt \sum_{n \in \{0, ..., N + 1\}} n \mu (f^{- 1} (\{n\})) \le \sum_{n \in \{0, ..., N + 1\}} n \mu ({f_{N + 1}}^{- 1} (\{n\}))\).
So, \(\int_M f d \mu = Sup (\{\int_M h d \mu \vert h \in P^+ \text{ such that } h \le f\}) = \infty\).
So, \(\int_M f d \mu = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} n \mu (f^{- 1} (\{n\}))\).
Let us suppose that \(\sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (f^{- 1} (\{n\})) = s \lt \infty\).
For each \(\epsilon \in \mathbb{R}\) such that \(0 \lt \epsilon\), there is an \(N \in \mathbb{N}\) such that \(\vert \sum_{n \in \{0, ..., N + 1\}} n \mu (f^{- 1} (\{n\})) - s \vert \lt \epsilon\), which means that \(s - \sum_{n \in \{0, ..., N + 1\}} n \mu (f^{- 1} (\{n\})) \lt \epsilon\).
\(s - \epsilon \lt \sum_{n \in \{0, ..., N + 1\}} n \mu (f^{- 1} (\{n\}))\).
Then, there is the non-negative simple measurable function, \(f_{N + 1}: M \to \mathbb{R}, m \mapsto Min (\{f (m), N + 1\})\), such that \(f_{N + 1} \le f\): which is indeed non-negative simple measurable, as before.
Then, \(\sum_{n \in \{0, ..., N + 1\}} n \mu (f^{- 1} (\{n\})) \le \sum_{n \in \{0, ..., N + 1\}} n \mu ({f_{N + 1}}^{- 1} (\{n\}))\), as before.
So, \(s - \epsilon \lt \sum_{n \in \{0, ..., N + 1\}} n \mu (f^{- 1} (\{n\})) \le \sum_{n \in \{0, ..., N + 1\}} n \mu ({f_{N + 1}}^{- 1} (\{n\}))\).
So, \(s = Sup (\{\int_M h d \mu \vert h \in P^+ \text{ such that } h \le f\}) = \int_M f d \mu\), by the proposition that for any linearly-ordered set and any subset, any element of the set is the supremum of the subset if and only if the element is equal to or larger than each element of the subset and for each element of the set smaller than the element, there is an element of the subset larger: while \(\int_M h d \mu \le \sum_{n \in \mathbb{N} \cup \{\infty\}} n \mu (f^{- 1} (\{n\})) = s\) has been seen above, \(s - \epsilon\) is "element of set smaller than the element" and \(\sum_{n \in \{0, ..., N + 1\}} n \mu ({f_{N + 1}}^{- 1} (\{n\}))\) is "element of the subset larger".
So, \(\int_M f d \mu = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} n \mu (f^{- 1} (\{n\}))\), anyway.
Step 2:
\(\sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} \mu (f^{-1} ([n, \infty])) = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} \mu (\cup_{j \in (\mathbb{R} \setminus \{0, ..., n - 1\}) \cup \{\infty\}} f^{-1} (\{j\})) = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} \sum_{j \in (\mathbb{R} \setminus \{0, ..., n - 1\}) \cup \{\infty\}} \mu (f^{-1} (\{j\})) = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} \sum_{j \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} s_{n, j}\) where \(s_{n, j} := 0\) when \(j \lt n\) and \(s_{n, j} := \mu (f^{-1} (\{j\}))\) when \(n \le j\), \(= \sum_{j \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} s_{n, j}\), by the proposition that for any absolutely convergent double series on the \(1\)-dimensional Euclidean metric space, the series with the sums orders changed converge to the same convergence and the proposition that for any divergent non-negative double series on the \(1\)-dimensional Euclidean metric space, the series with the sums orders changed diverges, \(= 1 \mu (f^{-1} (\{1\})) + 2 \mu (f^{-1} (\{2\})) + ... + \infty \mu (f^{-1} (\{\infty\})) = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} n \mu (f^{- 1} (\{n\}))\).
Step 3:
By Step 1 and Step 2, \(\int_M f d \mu = \sum_{n \in (\mathbb{N} \cup \{\infty\}) \setminus \{0\}} \mu (f^{-1} ([n, \infty]))\).
