Almost Everywhere and Almost Surely in Measure Theory 📂Measure Theory

Almost Everywhere and Almost Surely in Measure Theory

Definition ¹

When a function $f : E \to \overline{\mathbb{R}}$ has some property $P$ except on $E_{0} \subset E$ with $m(E_{0}) = 0$, we say that $f$ has $P$ almost everywhere on $E$.

Notation

When talking about probability, almost Everywhere is expressed as almost Surely, and when writing it out only in words is cumbersome, the following abbreviations are used. $$ f = g \text{ a.e.} \\ P(E) = 0 \text{ a.s.} $$

Explanation

Simply put, it means regarding every point except a null set as ‘almost everywhere’. This concept is only unfamiliar because it is newly defined; we already knew it when we learned the definite integral in high school. That is why, when the upper and lower limits were equal, the definite integral was necessarily $0$, and when computing probabilities we ignored whether the endpoints were included or not.

Basic Properties

[1]: If $f : E \to \mathbb{R}$ is measurable and $f = g$ almost everywhere on $E$, then $g$ is measurable on $E$.
[2]: If $f,g$ are measurable on $E$ and $|f| , |g| < \infty$ almost everywhere on $E$, then $\alpha f + \beta g$ is measurable on $E$.
[3]: If $f,g$ are measurable on $E$ and $|f| , |g| < \infty$ almost everywhere on $E$, then $f g$ is measurable.

Proof

It is good to prove these properties by hand at least once, but in terms of the fun of proving, there is not much worth seeing except [3].

[1]

Let $E_{0} = \left\{ x \in E \ | \ f(x) \ne g(x) \right\}$; then $E_{0} \subset E$ and $m(E_{0}) = 0$. For arbitrary $c$, $$ \left\{ x \in E \ | \ g(x) > c \right\} = \left\{ x \in E_{0} \ | \ g(x) > c \right\} \cup \left[ \left\{ x \in E \ | \ f(x) > c \right\} \cap ( E \setminus E_{0} ) \right] $$ Consider this. Examining the terms on the right-hand side one by one, since $\left\{ x \in E_{0} \ | \ g(x) > c \right\} \subset E_{0}$, $$ \left\{ x \in E_{0} \ | \ g(x) > c \right\} \in \mathcal{M} $$ Since $f$ is measurable on $E$, $$ \left\{ x \in E \ | \ f(x) > c \right\} \in \mathcal{M} $$ Finally, $$ E \cap (\mathbb{R} \setminus E_{0}) = ( E \setminus E_{0} ) \in \mathcal{M} $$ Therefore $\left\{ x \in E \ | \ g(x) > c \right\} \in \mathcal{M}$, and $g$ is measurable on $E$.

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[2]

If $\alpha = 0$ then $\alpha f$ is measurable, and if $\beta = 0$ then $\beta g $ is measurable.

If $\alpha \ne 0$, since $f$ is measurable, for arbitrary $\displaystyle {{c} \over {\alpha}}$, $$ \left\{ x \in E \ \left| \ f(x) > {{c} \over {\alpha}} \right. \right\} \in \mathcal{M} $$ Here, if $\alpha> 0$, $$ \left\{ x \in E \ | \ \alpha f(x) > c \right\} \in \mathcal{M} $$ and if $\alpha <0$, $$ \left\{ x \in E \ | \ \alpha f(x) < c \right\} \in \mathcal{M} $$ Therefore $\alpha f$ is measurable, and in the same way we can show that $\beta g$ is measurable when $\beta \ne 0$.

Now, if we show that $(f + g)$ is measurable, that is $\left\{ x \in E \ | \ f(x) + g(x) < c \right\} \in \mathcal{M}$, the proof is complete. Since the two functions take finite values, for every $x \in E$ there exists $c \in \mathbb{R}$ satisfying $f(x) + g(x) < c$. Rewriting, $f(x) < c - g(x)$, and by the density of the rationals there exists $q \in \mathbb{Q}$ satisfying $f(x) < q < c - g(x)$. Then $$ \bigcup_{q \in \mathbb{Q}} \left\{ x \in E \ | \ g(x) < c - q \right\} \cap \left\{ x \in \ | \ E f(x) < q \right\} = \left\{ x \in E \ | \ f(x) + g(x) < c \right\} \in \mathcal{M} $$

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Strategy[3]**: The idea of showing that $fg$ is measurable is summarized by the single identity $\displaystyle fg = {{1} \over {2}} \left[ (f+ g)^2 - f^2 - g^2\right]$.

[3]

Since we already showed in [2] that the sum of measurable functions whose values do not diverge is measurable, it suffices to show that $f^2$ is measurable. Since $f$ is measurable, for all $c$, $$ \left\{ x \in E \ | \ f(x) > \sqrt{c} \right\} \in \mathcal{M} \\ \left\{ x \in E \ | \ f(x) < - \sqrt{c} \right\} \in \mathcal{M} $$ Therefore $$ \left\{ x \in E \ | \ f(x) > \sqrt{c} \right\} \cup \left\{ x \in E \ | \ f(x) < - \sqrt{c} \right\} = \left\{ x \in E \ | \ f^2 (x) > c \right\} \in \mathcal{M} $$

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