The Set of p-Integrable and Essentially Bounded Functions

# The Set of p-Integrable and Essentially Bounded Functions

 Definition: Let $X$ be a set and let $\mathfrak T \subseteq \mathcal P(X)$. Then $(X, \mathfrak T)$ is called a Measurable Space. If $A \in \mathfrak T$ then $A$ is said to be a Measurable Set.
 Definition: Let $(X, \mathfrak T)$ be a measurable space. A function $\mu : \mathfrak T \to [0, \infty]$ is a Measure on $(X, \mathfrak T)$ if it satisfies the following properties: 1) $\mu (\emptyset) = 0$. 2) If $\{ A_n : n \in \mathbb{N} \} \subseteq \mathfrak T$ then $\displaystyle{\bigcup_{n=1}^{\infty} A_n \in \mathfrak T}$. (Countably Additivity) If $(X, \mathfrak T)$ is a measurable space and $\mu : \mathfrak T \to [0, \infty]$ is a measure, then the triple $(X, \mathfrak T, \mu)$ is said to be a Measure Space.
 Definition: Let $(X, \mathfrak T, \mu)$ be a measure space. A function $f : X \to \mathbb{C}$ is Measurable if for all $a \in \mathbb{C}$ and for all $\delta > 0$ we have that $f^{-1} ( \{ z \in \mathbb{C} : \mid z - a \mid < \delta \}) \in \mathfrak T$.

We are now ready to define the set of $p$-integrable and essentially bounded functions

 Definition: Let $0 < p \leq \infty$ and let $(X, \mathfrak T, \mu)$ be a measurable space. For $0 < p < \infty$, the Set of $p$-Integrable Functions is denoted $\mathcal L^p (X, \mathfrak T, \mu)$ and is the set of measurable functions $f : X \to \mathbb{C}$ such that $\displaystyle{\int_X |f|^p \: d \mu < \infty}$. We define the $p$-Seminorm on $\mathcal L^p (X, \mathfrak T, \mu)$ for all $f \in \mathcal L^p (X, \mathfrak T, \mu)$ by $\displaystyle{\| f \|_p = \left ( \int_X |f|^p \: d \mu \right )^{1/p}}$. For $p = \infty$, the Set of Essentially Bounded Functions is denoted $\mathcal L^{\infty} (X, \mathfrak T, \mu)$ and is the set of all measurable functions $f : X \to \mathbb{C}$ such that there exists an $M \in \mathbb{R}$ with $M > 0$ such that $|f(x)| \leq M$ $\mu$-almost everywhere on $X$. We define the $\infty$-Seminorm on $\mathcal L^{\infty} (X, \mathfrak T, \mu)$ for all $f \in \mathcal L^{\infty} (X, \mathfrak T, \mu)$ by $\displaystyle{\| f \|_{\infty} = \inf \left \{ M > 0 : | f(x) | \leq M \: \mu-\mathrm{a.e. \: on \:} X \right \}}$.

Recall that if $(X, \mathfrak T, \mu)$ is a measure space then a property $P$ is said to hold $mu$-almost everywhere or $\mu$-a.e. on $X$ if there exists a subset $X_0 \subseteq X$ such that the property $P$ holds on $X \setminus X_0$ and $P$ does not hold on $X_0$ where $\mu (X_0) = 0$. Thus, a measurable function $f : X \to \mathbb{C}$ is essentially bounded if it is bounded except on a set of measure $0$.

 Theorem 1: Let $0 < p \leq \infty$ and let $(X, \mathfrak T, \mu)$ be a measure space. a) If $\lambda \in \mathbb{C}$ and $f \in \mathcal L^p (X, \mathfrak T, \mu)$ then $\lambda f \in \mathcal L^p (X, \mathfrak T, \mu)$. b) If $p \geq 1$ and $f, g \in \mathcal L^p (X, \mathfrak T, \mu)$ then $(f + g) \in \mathcal L^p (X, \mathfrak T, \mu)$.
• Proof of a) There are two cases to consider.
• Case 1: Suppose that $0 < p < \infty$ and that $f \in \mathcal L^p (X, \mathfrak T, \mu)$. Then $\displaystyle{\int_X |f|^p \: d \mu < \infty}$ and:
(1)
\begin{align} \quad \int_X |\lambda f|^p \: d \mu = \int_X |\lambda|^p |f|^p \: d \mu = | \lambda |^p \int_X |f|^p \: d \mu < \infty \end{align}
• Therefore $\lambda f \in \mathcal L^p (X, \mathfrak T, \mu)$.
• Case 2: Suppose that $p = \infty$ and that $f \in \mathcal L^{\infty} (X, \mathfrak T, \mu)$. Then there exists an $M \in \mathbb{R}$, $M > 0$ such that $| f(x) | \leq M$ $\mu$-almost everywhere on $X$. Consider the function $\lambda f$. Then:
(2)
\begin{align} \quad |\lambda f(x)| = |\lambda||f(x)| \leq \lambda|M \quad \mu-\mathrm{almost \: everywhere \: on \:} X \end{align}
• Therefore $\lambda f \in \mathcal L^{\infty} (X, \mathfrak T, \mu)$. $\blacksquare$
• Proof of b) Once again, there are two cases to consider.
• Case 1: Suppose that $1 \leq p < \infty$ and that $f, g \in \mathcal L^p (X, \mathfrak T, \mu)$. We first note that the function $F(t) = t^p$ is a convex function on $[0, \infty)$, that is, for all $x, y \in [0, \infty)$ and for all $s \in [0, 1]$ we have that:
(3)
\begin{align} \quad F(sx + (1 - s)y) \leq sF(x) + (1 - s)F(y) \end{align}
• By setting $\displaystyle{s = \frac{1}{2}}$ we have that:
(4)
\begin{align} \quad \left ( \frac{x + y}{2} \right)^p \leq \frac{x^p}{2} + \frac{y^p}{2} \quad (*) \end{align}
• So for each point $x \in X$ we have that $|f(x)|, |g(x)| \geq 0$. Hence:
(5)
\begin{align} \quad \left ( \frac{|f(x) + g(x)|}{2} \right )^p \overset{\mathrm{Tri.}} \leq \left ( \frac{|f(x)| + |g(x)|}{2} \right )^p \overset{(*)} \leq \frac{|f(x)|^p}{2} + \frac{|g(x)|^p}{2} \end{align}
• Therefore we have that:
(6)
\begin{align} \quad (\| f + g \|_p)^p &= \left ( \left ( \int_X |f + g|^p \: d \mu \right )^{1/p} \right )^p \\ &= \int_X |f + g|^p \: d \mu \\ &= \int_X 2^p \left ( \frac{|f + g|}{2} \right )^p \: d \mu \\ &= 2^p \int_X \left ( \frac{|f + g|}{2} \right )^p \: d \mu \\ & \leq 2^p \int_X \left ( \frac{|f|^p}{2} + \frac{|g|^p}{2} \right ) \: d \mu \\ & \leq 2^p \int_X \frac{|f|^p}{2} \: d \mu + 2^p \int_X \frac{|g|^p}{2} \: d \mu \\ & \leq 2^{p-1} \int_X |f|^p \: d \mu + 2^{p-1} \int_X |g|^p \: d \mu \\ & \leq 2^{p-1} \| f \|^p + 2^{p-1} \| g \|^p \end{align}
• Since $f, g \in \mathcal L^p (X, \mathfrak T, \mu)$ we have that $\| f \|^p, \| g \|^p < \infty$. Therefore $\| f + g \|^p < \infty$. Hence $\displaystyle{\int_X |f + g|^p \: d \mu < \infty}$ which shows that $(f + g) \in \mathcal L^p (X, \mathfrak T, \mu)$.
• Case 2: Suppose that $p = \infty$ and that $f, g \in \mathcal L^{\infty} (X, \mathfrak T, \mu)$. Then there exists $M, N \in \mathbb{R}$ with $M, N > 0$ such that $|f(x)| \leq M$ $mu$-a.e. on $X$ and $|g(x)| \leq N$ $mu$-a.e. on $X$. So there exists sets $X_M, X_N \subseteq X$ such that $|f(x)| \leq M$ for all $x \in X \setminus X_M$ and $|g(x)| \leq N$ for all $x \in X \setminus X_N$ and $\mu (X_M) = 0$, $\mu (X_N) = 0$. Let $X_0 = X_M \cup X_n$. Then $|f(x)| + |g(x)| \leq M + N$ for all $x \in X \setminus X_0$ and $\mu (X_0) = 0$. Hence $(f + g) \in \mathcal L^{\infty} (X, \mathfrak T, \mu)$. $\blacksquare$
 Theorem 2: Let $1 \leq p \leq \infty$ and let $(X, \mathfrak T, \mu)$ be a measure space. Then $\mathcal L^p (X, \mathfrak T, \mu)$ is a vector space over $\mathbb{C}$ with the operations of function addition and scalar multiplication.
• Proof: The set $\mathcal L^p(X, \mathfrak T, \mu)$ is a set of measurable functions $f : X \to \mathbb{C}$. The basic vector space axioms are clearly satisfied. That is, for all $f, g, h \in \mathcal L^p (X, \mathfrak T, \mu)$ and for all $\lambda, \delta \in \mathbb{C}$ we have that $f + (g + h) = (f + g) + h$ (associativity of $+$), $f + g = g + f$ (commutativity of $+$), $0 \in \mathcal L^p (X, \mathfrak T, \mu)$ (existence of an additive identity), $-f \in \mathcal L^p (X, \mathfrak T, \mu)$ (existence of additive inverses), $\lambda (\delta f) = (\lambda \delta) f$ (associativity of scalar multiplication), $1f = f$ (existence of a scalar identity), and $\lambda (f + g) = \lambda f + \lambda g$, $(\lambda + \delta) f = \lambda f + \delta f$ (distributivity).
• By Theorem 1 we have that $\mathcal L^p (X, \mathfrak T, \mu)$ is closed under scalar multiplication and addition. So $\mathcal L^p(X, \mathfrak T, \mu)$ is a vector space. $\blacksquare$