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Tuesday, June 18, 2013

2.19

First, noting the definition of our set E_\alpha: E_\alpha = \big\lbrace x \in \mathbb{R}^d \hspace{0.25cm} \big| \hspace{0.25cm} |f(x)| > \alpha \big\rbrace Observe that E_\alpha is certainly measurable. Now, consider the integration: \int_0^\infty m(E_\alpha) \hspace{0.1cm} d\alpha = \int_0^\infty \int_{E_\alpha} dx \hspace{0.1cm} d\alpha Since constant functions over measurable sets are measurable, we can apply Tonelli's Theorem to see: \int_0^\infty \int_{E_\alpha} dx \hspace{0.1cm} d\alpha = \int_{E_\alpha} \int_0^\infty d\alpha \hspace{0.1cm} dx = \int_{\mathbb{R}^d} \int_0^\infty \chi_{[0,|f(x)|)} d\alpha \hspace{0.1cm} dx Next, notice from the definition of E_\alpha that: \int_{\mathbb{R}^d} \int_0^\infty \chi_{[0,|f(x)|)} d\alpha \hspace{0.1cm} dx = \int_{\mathbb{R}^d} m\big([0,|f(x)|)\big) \hspace{0.1cm} dx = \int_{\mathbb{R}^d} |f(x)| dx ...as desired.

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