description/proof of that for finite maps from set into Euclidean measurable space, \(\sigma\)-algebra induced on domain by sum of maps is contained in \(\sigma\)-algebra induced on domain by maps
Topics
About: measurable space
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean measurable space.
- The reader knows a definition of \(\sigma\)-algebra induced on domain of maps into measurable space.
Target Context
- The reader will have a description and a proof of the proposition that for any finite maps from any set into any Euclidean measurable space, the \(\sigma\)-algebra induced on the domain by the sum of the maps is contained in the \(\sigma\)-algebra induced on the domain by the maps.
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:
\(S_1\): \(\in \{\text{ the sets }\}\)
\(\mathbb{R}^d\): \(= \text{ the Euclidean measurable space }\)
\(J\): \(\in \{\text{ the finite index sets }\}\), such that \(J \neq \emptyset\)
\(\{f_j: S_1 \to \mathbb{R}^d \vert j \in J\}\):
\(\sum_{j \in J} f_j\): \(: S_1 \to \mathbb{R}^d\)
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Statements:
\(\sigma (\sum_{j \in J} f_j) \subseteq \sigma (\{f_j \vert j \in J\})\)
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2: Note
"\(\sigma (\sum_{j \in J} f_j) = \sigma (\{f_j \vert j \in J\})\)" does not necessarily hold.
For example, let \(S_1 = \mathbb{R}\), \(f_1: \mathbb{R} \to \mathbb{R}, r \mapsto r\), and \(f_2: \mathbb{R} \to \mathbb{R}, r \mapsto - r\). Then, \(\sum_{j \in J} f_j = 0\), so, \(\sigma (\sum_{j \in J} f_j) = \{\mathbb{R}, \emptyset\}\), but \(\{\mathbb{R}, \emptyset\} \subset \sigma (\{f_j \vert j \in J\})\), because \(\{\mathbb{R}, \emptyset\}\) does not make \(f_1\) measurable.
3: Proof
Whole Strategy: Step 1: see that with respect to \(\sigma (\{f_j \vert j \in J\})\), \(\sum_{j \in J} f_j\) is measurable.
Step 1:
With \(S_1\) given \(\sigma (\{f_j \vert j \in J\})\), \(f_j\) is measurable for each \(j \in J\), so, \(\sum_{j \in J} f_j\) is measurable, as is well known.
So, \(\sigma (\{f_j \vert j \in J\})\) is a \(\sigma\)-algebra that makes \(\sum_{j \in J} f_j\) measurable.
So, \(\sigma (\sum_{j \in J} f_j) \subseteq \sigma (\{f_j \vert j \in J\})\), because \(\sigma (\sum_{j \in J} f_j)\) is the smallest \(\sigma\)-algebra that makes \(\sum_{j \in J} f_j\) measurable, by Note for the definition of \(\sigma\)-algebra induced on domain of maps into measurable space.
