sums and products of measurable functions are measurable

[[concept]]

Topics

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Real-Valued Functions

Theorem

Let $E\subset \mathbb{R}$ be measurable and suppose $f,g:E\to \mathbb{R}$ are measurable functions, and $c\in \mathbb{R}$. Then $cf,f+g,$ and $fg$ are all measurable functions.

NOTE

this is useful because we will end up with $L^p$ spaces (L-p spaces) for Lebesgue integrable functions, which are often added together and multiplied.

Proof

(We just check the definition of measurable for each)

$cf$ is measurable If $c=0$ then $cf=0$ is a continuous function and thus measurable

• since continuous functions are measurable

So assume $c\ne 0$. If $\alpha \in \mathbb{R}$, then $cf(x)>\alpha ⟺f(x)>\frac{\alpha }{c}$ Thus we have $(cf{)}^{-1}((\alpha ,\infty ])=f^{-1}\left(\left(\frac{\alpha }{c},\infty \right]\right)$ and this is measurable for any $\alpha$ (since $f$ is measurable). Thus $cf$ is measurable.

$f+g$ is measurable If $\alpha \in \mathbb{R}$, then

f(x) + g(x) > \alpha &\iff f(c) >\alpha-g(x) \\ &\iff f(x) > r>\alpha - g(x) \end{align}$$ Where $r$ is some rational number (since the rationals are dense in the reals +++++). Thus there exists some $r \in \mathbb{Q}$ such that $$x \in f^{-1}((r, \infty]) \cap g^{-1}((\alpha-r, \infty])$$ And since both $f$ and $g$ are measurable, this intersection is also measurable. Thus the preimage is given as $$\begin{align} (f+g)^{-1}((\alpha, \infty]) &= \bigcup_{r \in \mathbb{Q}} (f^{-1} (\,(r, \infty]\,)) \cap g^{-1} (\,(\alpha-r, \infty]\,) \end{align}$$ Which is measurable, since this is a countable union. Thus $f+g$ is [[Concept Wiki/measurable function\|measurable]]

$fg$ is measurable We first show $f^2$ is measurable. Since $f^2$ is a nonnegative function, then for any $\alpha <0$ we have $(f^2{)}^{-1}((\alpha ,\infty ])=E$ because the entire domain maps within $(\alpha ,\infty ]$ and so this is measurable by assumption. If instead $\alpha \ge 0$, we have

f^2 > \alpha &\iff f(x) > \sqrt{ \alpha } \text{ or } f(x) < - \sqrt{ \alpha } \\ \implies (f^2)^{-1} ((\alpha, \infty]) &= f^{-1}((\sqrt{ \alpha }, \infty]) \;\cup \; f^{-1}([-\infty,-\sqrt{ \alpha })) \end{align}$$ And both sets on RHS are measurable, so using the [[Concept Wiki/equivalent intervals of measurability for measurable functions]], we see that the union is measurable and thus $f^2$ is measurable. Finally, we notice that $$fg = \frac{1}{4}(\,(f+g)^2 - (f-g)^2\,)$$ - And both $f+g$ and $f-g$ are measurable by the previous two proofs above.

\tag*{$\blacksquare$}

Complex-Valued Functions

Theorem

If $f,g:E\to \mathbb{C}$ are measurable and $\alpha \in \mathbb{C}$, then

• $\alpha f$
• $f+g$
• $fg$
• $\bar{f}$
• $|f|$

Are all measurable as well.

Proof

Exercise

References

References

See Also

Mentions

Mentions

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