description/proof of that for measure space and measurable subset, Lebesgue integral of measurable extended real function over space is integral over subset plus integral over complement of subset
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 the real numbers field with the canonical linear ordering and any finite set of subsets, if the sum of each set of representatives of the subsets is equal to or smaller than any number, the sum of the supremums of the subsets is equal to or smaller than the number.
Target Context
- The reader will have a description and a proof of the proposition that for any measure space and for any measurable subset, the Lebesgue integral of any measurable extended real function over the space is the integral over the subset plus the integral over the complement of the subset.
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:
\((M, A, \mu)\): \(\in \{\text{ the measure spaces }\}\)
\(\overline{\mathbb{R}}\): \(= \text{ the extended Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(\mathbb{R}\): \(= \text{ the Euclidean topological space }\) with the Borel \(\sigma\)-algebra
\(f\): \(: M \to \overline{\mathbb{R}}\), \(\in \{\text{ the measurable maps }\}\)
\(a\): \(\in A\)
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Statements:
\(\int_M f d \mu = \int_a f d \mu + \int_{M \setminus a} f d \mu\)
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2: Proof
Whole Strategy: Step 1: see what \(\int_M f d \mu\), \(\int_a f d \mu\), and \(\int_{M \setminus a} f d \mu\) mean; Step 2: see that for each \(g = f^+ \text{ or } f^-\), \(\int_a g d \mu + \int_{M \setminus a} g d \mu \le \int_M g d \mu\); Step 3: see that for each \(g = f^+ \text{ or } f^-\), \(\int_M g d \mu \le \int_a g d \mu + \int_{M \setminus a} f g \mu\); Step 4: conclude the proposition.
Step 1:
\(\int_M f d \mu = \int_M f^+ d \mu - \int_M f^- d \mu\), \(\int_a f d \mu = \int_M \chi_a f^+ d \mu - \int_M \chi_a f^- d \mu\), and \(\int_{M \setminus a} f d \mu = \int_M \chi_{M \setminus a} f^+ d \mu - \int_M \chi_{M \setminus a} f^- d \mu\), by definition.
Let \(g := f^+ \text{ or } f^-\).
\(\int_M g d \mu = Sup (\{\int_M h d \mu \vert h \in P^+ \text{ such that } h \le g\})\), where \(\int_M h d \mu\) means for \(h (M) = \{t_1, ..., t_n\}\) (which means that \(h = \sum_{j \in \{1, ..., n\}} t_j \chi_{h^{-1} (t_j)}\)), \(\sum_{j \in \{1, ..., n\}} t_j \mu (h^{-1} (t_j))\), by definition.
\(\{h^{-1} (t_j) \vert j \in \{1, ..., n\}\}\) are some disjoint measurable subsets of \(M\).
\(\int_a g d \mu = Sup (\{\int_M h d \mu \vert h \in P^+ \text{ such that } h \le \chi_a g\})\), where \(h = \sum_{j \in \{1, ..., n_a\}} r_j \chi_{h^{-1} (r_j)}\) and \(\int_M h d \mu = \sum_{j \in \{1, ..., n_a\}} r_j \mu (h^{-1} (r_j))\).
\(\{h^{-1} (r_j) \vert j \in \{1, ..., n_a\}\}\) are some disjoint measurable subsets of \(a\).
\(\int_{M \setminus a} g d \mu = Sup (\{\int_M h d \mu \vert h \in P^+ \text{ such that } h \le \chi_{M \setminus a} g\})\), where \(h = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \chi_{h^{-1} (s_j)}\) and \(\int_M h d \mu = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \mu (h^{-1} (s_j))\).
\(\{h^{-1} (s_j) \vert j \in \{1, ..., n_{M \setminus a}\}\}\) are some disjoint measurable subsets of \(M \setminus a\).
Step 2:
Let us see that \(\int_a g d \mu + \int_{M \setminus a} g d \mu \le \int_M g d \mu\).
Let \(h_a = \sum_{j \in \{1, ..., n_a\}} r_j \chi_{{h_a}^{-1} (r_j)} \in \{h \in P^+ \text{ such that } h \le \chi_a g\}\) and \(h_{M \setminus a} = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \chi_{{h_{M \setminus a}}^{-1} (s_j)} \in \{h \in P^+ \text{ such that } h \le \chi_{M \setminus a} g\}\}\) be any.
\(h_a + h_{M \setminus a} \in \{h \in P^+ \text{ such that } h \le g\}\), because it is a simple function with the finite values, \(\{r_1, ..., r_{n_a}, s_1, ..., s_{n_{M \setminus a}}\}\) (which may contain some duplications), where \(\{(h_a + h_{M \setminus a})^{-1} (r_1), ..., (h_a + h_{M \setminus a})^{-1} (r_{n_a}), (h_a + h_{M \setminus a})^{-1} (s_1), ..., (h_a + h_{M \setminus a})^{-1} (s_{n_{M \setminus a}})\}\) (which may contain some duplications) are some disjoint measurable subsets of \(M\): \((h_a + h_{M \setminus a})^{-1} (r_1) = {h_a}^{-1} (r_1) \cup {h_{M \setminus a}}^{-1} (r_1)\) where \({h_{M \setminus a}}^{-1} (r_1)\) is for when \(r_1 \in \{s_1, ..., s_{n_{M \setminus a}}\}\), and so on, and \(h_a + h_{M \setminus a} \le g\) holds over \(a\), because \(h_a + h_{M \setminus a} = h_a + 0 \le \chi_a g = g\), and it holds over \(M \setminus a\), because \(h_a + h_{M \setminus a} = 0 + h_{M \setminus a} \le \chi_{M \setminus a} g = g\), so it holds over \(M\).
So, \(\int_M h_a d \mu + \int_M h_{M \setminus a} d \mu = \int_M (h_a + h_{M \setminus a}) d \mu \le \int_M g d \mu\).
By the proposition that for the real numbers field with the canonical linear ordering and any finite set of subsets, if the sum of each set of representatives of the subsets is equal to or smaller than any number, the sum of the supremums of the subsets is equal to or smaller than the number, \(\int_a g d \mu + \int_{M \setminus a} g d \mu = Sup (\{\int_M h_a d \mu\}) + Sup (\{\int_M h_{M \setminus a} d \mu\}) \le \int_M g d \mu\).
Step 3:
Let us see that \(\int_M g d \mu \le \int_a g d \mu + \int_{M \setminus a} g d \mu\).
Let \(h = \sum_{j \in \{1, ..., n\}} t_j \chi_{h^{-1} (t_j)} \in \{h \in P^+ \text{ such that } h \le g\}\) be any.
Let us take \(\{r_1, ..., r_{n_a}\} = \{r_j \in \{t_1, ..., t_n\} \vert h^{-1} (t_1) \cap a \neq \emptyset\}\), which may be empty, and \(\{s_1, ..., s_{n_{M \setminus a}}\} = \{s_j \in \{t_1, ..., t_n\} \vert h^{-1} (t_1) \cap (M \setminus a) \neq \emptyset\}\), which may be empty.
Let us take \(h_a = \sum_{j \in \{1, ..., n_a\}} r_j \chi_{h^{-1} (r_j) \cap a}\), which is regarded to be \(0\) when \(\{1, ..., n_a\}\) is empty and \(h_{M \setminus a} = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \chi_{h^{-1} (s_j) \cap (M \setminus a)}\), which is regarded to be \(0\) when \(\{1, ..., n_{M \setminus a}\}\) is empty.
Then, \(h = h_a + h_{M \setminus a}\), because over \(a\), \(h = \sum_{j \in \{1, ..., n\}} t_j \chi_{h^{-1} (t_j)} = \sum_{j \in \{1, ..., n_a\}} r_j \chi_{h^{-1} (r_j) \cap a}\), because only the terms with \(h^{-1} (t_j) \cap a \neq \emptyset\) do not vanish, \(= h_a = h_a + h_{M \setminus a}\), and over \(M \setminus a\), \(h = \sum_{j \in \{1, ..., n\}} t_j \chi_{h^{-1} (t_j)} = \sum_{j \in \{1, ..., n_{M \setminus a}\}} s_j \chi_{h^{-1} (s_j) \cap (M \setminus a)}\), because only the terms with \(h^{-1} (t_j) \cap (M \setminus a) \neq \emptyset\) do not vanish, \(= h_{M \setminus a} = h_a + h_{M \setminus a}\).
\(h_a \in \{h \in P^+ \text{ such that } h \le \chi_a g\}\), because it is a simple function with the finite values, \(\{r_1, ..., r_{n_a}\}\), where \(\{{h_a}^{-1} (r_1), ..., {h_a}^{-1} (r_{n_a})\}\) are some disjoint measurable subsets of \(M\): \({h_a}^{-1} (r_1) = h^{-1} (r_1) \cap a\), and so on, and \(h_a \le \chi_a g\) holds over \(a\), because \(h_a = h \le g = \chi_a g\), and it holds over \(M \setminus a\), because \(h_a = 0 \le \chi_a g\), so it holds over \(M\).
\(h_{M \setminus a} \in \{h \in P^+ \text{ such that } h \le \chi_{M \setminus a} g\}\), because it is a simple function with the finite values, \(\{s_1, ..., s_{n_{M \setminus a}}\}\), where \(\{{h_{M \setminus a}}^{-1} (s_1), ..., {h_{M \setminus a}}^{-1} (s_{n_{M \setminus a}})\}\) are some disjoint measurable subsets of \(M\): \({h_{M \setminus a}}^{-1} (s_1) = h^{-1} (s_1) \cap (M \setminus a)\), and so on, and \(h_{M \setminus a} \le \chi_{M \setminus a} g\) holds over \(a\), because \(h_{M \setminus a} = 0 \le \chi_{M \setminus a} g\), and it holds over \(M \setminus a\), because \(h_{M \setminus a} = h \le g = \chi_{M \setminus a} g\), so it holds over \(M\).
\(\int_M h d \mu = \int_M (h_a + h_{M \setminus a}) d \mu = \int_M h_a d \mu + \int_M h_{M \setminus a} d \mu \le \int_a g d \mu + \int_{M \setminus a} g d \mu\).
That implies that \(\int_M g d \mu \le \int_a g d \mu + \int_{M \setminus a} g d \mu\).
Step 4:
So, \(\int_M g d \mu = \int_a g d \mu + \int_{M \setminus a} g d \mu\).
So, \(\int_M f d \mu = \int_M f^+ d \mu - \int_M f^- d \mu = \int_a f^+ d \mu + \int_{M \setminus a} f^+ d \mu - (\int_a f^- d \mu + \int_{M \setminus a} f^- d \mu) = \int_a f^+ d \mu - \int_a f^- d \mu + \int_{M \setminus a} f^+ d \mu - \int_{M \setminus a} f^- d \mu = \int_a (f^+ - f^-) d \mu + \int_{M \setminus a} (f^+ - f^-) d \mu = \int_a f d \mu + \int_{M \setminus a} f d \mu\).
