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Suppose that $X$ is a subset of a Hausdorff topological space, and $Z\subseteq X$ a member of the family of Borel subsets on $X$. Let $B(Z)$ the set of bounded functions on $Z$, equipped with the $\sup$ norm. Then we can easily prove that every Cauchy sequence in $B(Z)$ converges.

Now take $BM(Z)$ the set of real-valued, bounded, measurable functions defined on $Z$. The fact that the functions are measurable, gives a hint as to define a norm on this space that uses this measure. Indeed, suppose that $\mu$ such a (finite, signed) measure on $Z$; the essential supremum norm $\|f\|^*_\infty = \mathrm{ess}\sup |f|$ yields the equivalence class of measurable functions that are bounded $\mu-$almost everywhere. Call this space $\mathcal{L}_\infty(Z,B(Z),\mu)$.

Question:

Suppose that we take a function $f\in \mathcal{L}_\infty$. Then this function may or may not in general exist in $B(Z)$, for it may assume an infinite value at some point in $Z$. Then how can one be assured that $\mathcal{L}_\infty$ is a vector space?

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    Let me point out that $\mathcal{L}^\infty$ is not a good symbol: it is commonly used in the theory of Banach lattices (in particular, $c_0$ is an abstract $\mathcal{L}^\infty$ space. It's better to write simply $L^\infty$ to avoid confusion.2011-08-26

2 Answers 2

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The essential supremum norm ... yields the equivalence class of measurable functions that are bounded μ−almost everywhere.

This sentence indicates confusion. Each element of $\mathcal{L}_\infty$ is an equivalence class of measurable functions. Once you are used to this, you can go back to abusing notation and terminology like everyone else, but for now let's be careful. If $f$ is an essentially bounded function, let $[f]\in\mathcal{L}_\infty$ denote the equivalence class of $f$. If $[f]$ and $[g]$ are in $\mathcal{L}_\infty$, define $[f]+[g]=[f+g]$. If $[f]$ is in $\mathcal{L}_\infty$ and $a$ is in $\mathbb R$, define $a\cdot [f]=[af]$.

With these operations, $\mathcal{L}_\infty$ is a vector space. Checking that the operations satisfy the axioms is straightforward, but first you need to know that the definitions of the operations make sense. For example, to show that $+$ is "well-defined", you need to show:

  • If $f$ is essentially bounded and $g$ is essentially bounded, then $f+g$ is essentially bounded.
  • If $f_1=f_2$ a.e. and $g_1=g_2$ a.e., then $f_1+g_1=f_2+g_2$ a.e.

Both facts rely on the fact that a union of two sets of measure zero is a set of measure zero. Scalar multiplication is similar but easier.

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    Dealing with the infinities in measure theory always bothered me. One probably *has* to live with the fact that at some point one needs to deal with the "space" $\mathscr{L}^0$ of functions that are only defined (and measurable) on a co-null set. This isn't a vector space but as soon as one quotients out the functions that are $0$ a.e. one ends up with a vector space $L^0$. Then the $\mathscr{L}^p$ sub"space" of $\mathscr{L}^0$ of functions for which $|f|^p$ is integrable again suffers from the same defect but the Hausdorff quotient $L^p$ is a subspace of $L^0$, etc.2011-08-26
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Let $\mu$ be a positive measure defined on a $\sigma$-algebra $\mathcal{A}$ of subsets of the set $X$. It is easy to prove that the set $\mathcal{M}(X, \mathcal{A}, \mu)$ of all $\mu$- measurable functions $f \colon X \rightarrow \mathbb{K} \in \{ \mathbb{R}, \mathbb{C}\}$ is a linear subspace of the space of all maps from $X$ to $\mathbb{K}$.

Moreover, we can prove that the space $\mathcal{L}_{\infty}(X, \mathcal{A}, \mu)$ of all essentialy bounded functions $f \colon X \rightarrow \mathbb{K}$ is a linear subspace of $\mathcal{M}(X, \mathcal{A}, \mu)$. Indeed, let $f_1, \ f_2 \in \mathcal{L}_{\infty}$ and let $M_1$ and $M_2$ be such positive numbers that $|f_1(x)| \leq M_1$ and $|f_2(x)| \leq M_2$ almost every where on $X$. To show that $f_1+f_2 \in \mathcal{L}_{\infty}$ it is enough to note that $\{ x \in X \colon |f_1(x) + f_2(x)| > M_1+M_2\} \subset \{ x \in X \colon |f_1(x)| >M_1\} \cup \{ x \in X \colon |f_2(x)| >M_2 \},$ then $\mu (\{ x \in X \colon |f_1(x) +f_2(x) | > M_1+M_2\})=0.$ Verification that $\lambda f \in \mathcal{L}_{\infty}$ is even easier.