equivalent intervals of measurability for measurable functions

[[concept]]

Topics

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Theorem

Let $E\subset \mathbb{R}$ be measurable and let $f:E\to [-\infty ,+\infty ]$. Then the following are equivalent for all $\alpha \in \mathbb{R}$,

1. $f^{-1}((\alpha ,\infty ])\in \mathcal{M}$
2. $f^{-1}([\alpha ,\infty ])\in \mathcal{M}$
3. $f^{-1}([-\infty ,\alpha ))\in \mathcal{M}$
4. $f^{-1}([-\infty ,\alpha ])\in \mathcal{M}$

Proof

Not hard!

$(1)⟹(2)$ Suppose $(1)$ holds. Then for all $\alpha \in \mathbb{R}$

[\alpha, \infty] &= \bigcap_{n} \left( \alpha - \frac{1}{n}, \infty \right] \\ \implies f^{-1}(\,[\alpha, \infty]\,) &= \bigcap_{n} f^{-1} \left( \left( \alpha-\frac{1}{n}, \infty \right] \right) \end{align}$$ RHS is a countable intersection of measurable sets and is thus measurable.

$(2)⟹(1)$ Suppose $(2)$ holds. Then for all $\alpha \in \mathbb{R}$, $(\alpha ,\infty ]={\bigcup }_n\left[\alpha +\frac{1}{n},\infty \right]⟹f^{-1}(\alpha ,\infty ]={\bigcup }_n{f}^{-1}\left(\left[\alpha +\frac{1}{n},\infty \right]\right)$ And this is a countable union of measurable sets and is thus Lebesgue measurable

ie $(1)⟺(2)$

$(3)⟺(4)$ uses the same argument.

$(2)⟺(3)$ Note that $[-\infty ,\alpha )=([\alpha ,\infty ]{)}^c$. Thus by taking the preimages and noting that complements of measurable sets are measurable yields the desired result.

\tag*{$\blacksquare$}

References

References

See Also

Mentions

Mentions

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