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## Proof: (related to "Step Functions as a Subspace of all Functions on a Closed Real Interval")

According to the definition of a subspace, we have to verify the following properties:

1. $0\in T[a,b].$

1. If $\phi,\psi\in T[a,b]$ then $\phi+\psi\in T[a,b].$
• Let $\phi$ be a step function defined with respect to the partition $$a=x_0 < x_1 < \ldots < x_{n-1} < x_n=b$$ and let $\psi$ be a step function defined with respect to the partition $$a=y_0 < y_1 < \ldots < y_{m-1} < y_m=b.$$
• We define the set union of both partions:
$$\{t_0,t_1\ldots t_k\}:=\{x_0, x_1, \ldots, x_n\}\cup\{y_0, y_1, \ldots, y_m\},$$
and observe that $\phi$ and $\psi$ are both constant functions on every real interval $]t_{i-1},t_{i}[$, $i=1,\ldots,k.$
• Thus, $\phi+\psi$ is a step function and therefore $\phi+\psi\in T[a,b].$
1. If $\phi\in T[a,b]$ and $\lambda\in\mathbb R$ then $\lambda\phi\in T[a,b].$
• Obviously, if $\phi$ is a step function, so is $\lambda\phi.$
q.e.d

| | | | created: 2020-01-09 06:43:22 | modified: 2020-01-09 06:43:22 | by: bookofproofs | references: [581]

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