Integral inequality in a probabilty space

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Let $(X,\mathcal{A},\mu)$ be a probability space.

Let $v \in \mathcal{M}^+(\mathcal{A})$.

Show that $|\int_X v \, d\mu | \leq ||v||$.

$||v||$ denotes the uniform norm.

I know the following but would like some help to connect it with the uniform norm given by $ ||v|| = \sup \lbrace |v(x)| : x \in X \rbrace $

$|\int_X v \, d\mu | \leq \int_X |v| \, d\mu $.

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With $\|v\| =\sup \lbrace |v(x)| : x \in X \rbrace,$ calculate

$|\int_X v \, d\mu | \leq \int_X|v|d\mu\le \int_X\|v\| d\mu=$

$\|v\|\int_Xd\mu=\|v\|\cdot \mu (X)=\|v\|\cdot 1=\|v\|.$