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Chapter 6: Problem 4

Lemma. Let \(\mathscr{S}\) be a \(\sigma\)-algebra of subsets of a set \(X\), and denote by \(\mathscr{F}\) the class of functions \(f\) on \(X\) such that \(\\{f>t\\} \in \mathscr{S}\) for each \(t\) in \(\mathbb{R}\). Then \(\mathscr{F}\) is a sequentially complete, unital algebra of functions on \(X\), stable under \(\bigvee\) and \(\bigwedge\) Moreover, \(|f|^{p} \in \mathscr{F}\) for every \(f\) in \(\mathscr{F}\) and \(p>0\). Proof. If \(f\) and \(g\) belong to \(\mathscr{F}\), then $$ \\{f+g>t\\}=\bigcup\left(\left\\{f>r_{n}\right\\} \cap\left\\{g>t-r_{n}\right\\}\right) \in \mathscr{S}, $$ for each \(t\) in \(\mathbb{R}\), if \(\left(r_{n}\right)\) is an enumeration of the rational numbers. Thus, \(f+g \in \mathscr{F} .\) If \(\alpha>0\), then \(\\{\alpha f>t\\}=\left\\{f<\alpha^{-1} t\right\\} \in \mathscr{S}\), so that \(\alpha f \in \mathscr{F}\). Moreover,$$ \begin{aligned} \\{-f>t\\} &=\\{f<-t\\} \\ &=X \backslash\\{f \geq-t\\}=X \backslash \cap\left\\{f>-t-n^{-1}\right\\} \in \mathscr{S}, \end{aligned} $$ so that \(-f \in \mathscr{F}\). Taken together it shows that \(\mathscr{F}\) is a vector space. The expressions $$ \begin{aligned} &\\{f \vee g>t\\}=\\{f>t\\} \cup\\{g>t\\} \\ &\\{f \wedge g>t\\}=\\{f>t\\} \cap\\{g>t\\} \end{aligned} $$ show that \(\mathscr{F}\) is stable under the lattice operations \(\bigvee\) and \(\bigwedge\). If \(f \in \mathscr{F}\) and \(t \geq 0\), then \(\left\\{|f|^{P}>t\right\\}=\left\\{|f|>t^{1 / D}\right\\} \in \mathscr{S}\), because \(|f|=\) \(f \vee 0-f \wedge 0 \in \mathscr{F}\). If \(t<0\), then \(\left\\{|f|^{p}>t\right\\}=X \in \mathscr{S}\). It follows that \(|f|^{p} \in \mathscr{F}\) for every \(p>0\). As $$ f g=\frac{1}{2}\left((f+g)^{2}-f^{2}-g^{2}\right) $$ this implies (with \(p=2\) ) that \(f g \in \mathscr{F}\) for all \(f\) and \(g\) in \(\mathscr{F}\), so that \(\mathscr{F}\) is an algebra. To show that \(\mathscr{\text { is sequentially complete, it suffices by }} 6.2 .2\) to show that IF is monotone sequentially complete. Therefore, take a monotone increasing sequence \(\left(f_{n}\right)\) in \(\mathscr{F}\) that converges pointwise to some function \(f\). Then for each \(t\) in \(\mathbb{R}\) $$ \\{f>t\\}=\bigcup\left\\{f_{n}>t\right\\} \in \mathscr{S}, $$ so that \(f \in \mathscr{F}\), completing the proof.

Short Answer

Expert verified
\(\mathscr{F}\) is a sequentially complete, unital algebra, closed under addition, scalar multiplication, supremum, infimum, and multiplication.

Step by step solution

01

Understanding the Setup

We begin with understanding the context. We have a set \(X\) and a \(\sigma\)-algebra \(\mathscr{S}\) which is a collection of subsets of \(X\). The class \(\mathscr{F}\) contains functions \(f\) such that the set \(\{f > t\}\) is in \(\mathscr{S}\) for all real numbers \(t\). Our goal is to prove several properties of \(\mathscr{F}\).
02

Vector Space Properties

To prove that \(\mathscr{F}\) is a vector space, we show it is closed under addition and scalar multiplication. For any two functions \(f, g \in \mathscr{F}\), we see that \(\{f+g > t\} = \bigcup \left(\{f > r_n\} \cap \{g > t - r_n\}\right) \in \mathscr{S}\) for rational numbers \(r_n\). Similarly, for \(\alpha > 0\), \(\{\alpha f > t\} = \{f > \alpha^{-1}t\} \) shows closure under scalar multiplication. This establishes \(\mathscr{F}\) as a vector space.
03

Lattice Operations

Next, we prove stability under lattice operations \(\bigvee\) (supremum) and \(\bigwedge\) (infimum). For \(\vee\), \(\{f \vee g > t\} = \{f > t\} \cup \{g > t\}\), and for \(\wedge\), \(\{f \wedge g > t\} = \{f > t \} \cap \{g > t \}\), confirming \(\{\cdot > t\}\) remains in \(\mathscr{S}\). Thus, \(\mathscr{F}\) is closed under these operations.
04

Functional Powers

We show \( |f|^p \in \mathscr{F}\) for any \(p > 0\). For non-negative \(t\), \(\{|f|^p > t\} = \{|f| > t^{1/p}\} \in \mathscr{S}\). If \(t < 0\), then \(\{|f|^p > t\} \) is clearly just \(X\), also in \(\mathscr{S}\). This inclusion means \(|f|^p\) belongs in \(\mathscr{F}\).
05

Product Closure

To prove closure of \(\mathscr{F}\) under multiplication, note \(f g = \frac{1}{2}((f+g)^2 - f^2 - g^2)\). This representation shows multiplication can be expressed in terms of addition and squaring (with \(p = 2\)), thus \(f g \in \mathscr{F}\). Hence, \(\mathscr{F}\) is an algebra.
06

Sequential Completeness

Finally, we establish that \(\mathscr{F}\) is sequentially complete. For a monotone increasing sequence \( f_n \) converging to \(f\) and given \(\{f > t\} = \bigcup \{f_n > t\}\), and since each \(\{f_n > t\} \in \mathscr{S}\), \(\{f > t\} \in \mathscr{S}\) by the property of \(\sigma\)-algebras. Thus, \(f \in \mathscr{F}\), proving sequential completeness.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sigma-algebra
In mathematics, particularly in the field of measure theory, a sigma-algebra (\(\sigma\)-algebra) is a collection of subsets of a given set that not only includes the empty set but is also closed under complements and countable unions.
This means that if you have a sigma-algebra \(\mathscr{S}\) on a set \(X\), and you take any subset \(A\) of \(X\) from \(\mathscr{S}\), the complement of \(A\) with respect to \(X\) will also be in \(\mathscr{S}\).
Additionally, if \(A_1, A_2, A_3, \ldots\) are subsets in \(\mathscr{S}\), the union \(A_1 \cup A_2 \cup A_3 \cup \ldots\) will also be in \(\mathscr{S}\). This property is pivotal in defining measures which can talk about sizes or probabilities of more complicated sets by starting with basic, simpler elements.
  • Contains the whole set \(X\) and the empty set \(\emptyset\).
  • Closed under complements: If set is included, so is its complement.
  • Closed under countable unions: If a sequence of sets is included, so is their union.
Understanding sigma-algebras is essential because they provide the framework within which functions behave predictably with respect to measures, as seen in this exercise where function classifications depend directly on these foundational structures.
Vector spaces
A vector space is a fundamental concept in mathematics that provides a framework for constructing linear spaces. It is defined over a field, such as the real or complex numbers, and includes operations like addition of vectors and multiplication by scalars. In this context, the family of functions \(\mathscr{F}\) forms a vector space because it is closed under addition and scalar multiplication.
For \(\mathscr{F}\), closure under addition means if \(f\) and \(g\) belong to \(\mathscr{F}\), their sum \(f + g\) will also remain in \(\mathscr{F}\).
Closure under scalar multiplication implies that if \(\alpha\) is a scalar, and \(f\) belongs to \(\mathscr{F}\), then \(\alpha f\) is also in \(\mathscr{F}\).
This highlight on vector spaces is crucial for understanding how function spaces can be manipulated in a controlled manner using these operations.
  • Closed under addition: Sum of functions is still part of \(\mathscr{F}\).
  • Closed under scalar multiplication: Scaling a function keeps it within the space.
These properties make vector spaces incredibly useful in numerous applications, from solving linear equations to understanding function spaces like those being analyzed here.
Sequential completeness
Sequential completeness is a property of some mathematical structures and is indispensable in analysis. It signifies that in a given set, every Cauchy sequence converges to a limit within that set. In this exercise, \(\mathscr{F}\) being sequentially complete implies that if you have a sequence of functions \((f_n)\) in \(\mathscr{F}\) that converges, the limit function \(f\) will also belong to \(\mathscr{F}\).
For \(\mathscr{F}\) to be sequentially complete, if you consider a monotone increasing sequence of functions that converge pointwise to some function, the resulting limit function will still behave as required by the elements of \(\mathscr{F}\). This ensures that operation limits can be reliably handled without leaving the function space.
  • Convergence: Any convergent sequence of functions in \(\mathscr{F}\) has its limit in \(\mathscr{F}\).
  • Enables predictable behavior for sequences of functions.
Understanding sequential completeness gives assurance that sequences and limits in these controlled settings behave predictably, maintaining the algebraic structure necessary to advance in functional analysis.
Lattice operations
Lattice operations in the context of function spaces refer to the operations of infimum and supremum, commonly denoted as \(\wedge\) and \(\vee\). These operations are crucial when examining order relations within sets of functions.
For a collection of functions to be stable under lattice operations, it means that taking the least upper bound or greatest lower bound of two functions will yield another function within the same class. Here, \(\mathscr{F}\) is stable under lattice operations like \(f \vee g\) and \(f \wedge g\) because:
  • \(\{f \vee g > t\} = \{f > t\} \cup \{g > t\}\), showing the supremum is covered.
  • \(\{f \wedge g > t\} = \{f > t\} \cap \{g > t\}\), showing the infimum is covered.
These operations ensure that when you combine functions via these order-preserving ways, the resulting functions do not exit the theoretical framework of \(\mathscr{F}\).Understanding lattice operations helps recognize how function spaces are organized and how complex functions can be constructed through simpler function pairs with respect to their orders or magnitudes.

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Most popular questions from this chapter

Lemma. For a real function \(f\) on \(X \times Y\) and \(y\) in \(Y\), consider the function \(f(\cdot, y)\) on \(X .\) If \(f \in C_{c}(X \times Y)\), then \(f(\cdot, y) \in C_{c}(X) .\) If \(f \in C_{c}(X \times Y)^{m}\), then \(f(\cdot, y) \in C_{c}(X)^{m} .\) If \(f \in \mathscr{B}(X \times Y)\), then \(f(\cdot, y) \in \mathscr{B}(X)\). Similar results hold for each function \(f\left(x,^{\circ}\right), x \in X\). Proof. Let \(t_{y}: X \rightarrow X \times Y\) be the continuous function given by \(l_{y}(x)=(x, y)\). Then \(f(\cdot, y)=f \circ t_{y}\). From this we see immediately that if \(f \in C_{c}(X \times Y)\), then \(f(\cdot, y)\) is continuous. Furthermore, it is clear that the support of \(f(\cdot, y)\) is contained in the projection from \(X \times Y\) onto \(X\) of the support of \(f\), so that \(f(\cdot, y) \in C_{c}(X)\). It follows from the definitions that \(f(\cdot, y) \in C_{c}(X)^{m}\) if \(f \in\) \(C_{c}(X \times Y)^{m}\). Finally, since composition of Borel functions again produce Borel functions, we see from the identity \(f(\cdot, y)=f \circ t_{y}\) that \(f(\cdot, y) \in \mathscr{B}(X)\) if \(f \in \mathscr{B}(X \times Y)\).

Theorem. For a locally compact group \(G\) with Haar integral \(\int\), the space \(L^{1}(G)\) is a Banach algebra with an isometric involution. Product and involution in \(L^{1}(G)\) are given by the formulas $$ f \times g(x)=\int f(y) g\left(y^{-1} x\right) d y, \quad f^{*}(x)=\overline{f\left(x^{-1}\right)} \Delta(x)^{-1} $$ Proof. It follows from \(6.6 .20\) that \(f \times g \in L^{1}(G)\) with \(\|f \times g\|_{1} \leq\|f\|_{1}\|g\|_{1}\). It is elementary to check that the convolution product is distributive with respect to the sum, but it requires Fubini's theorem to show that the product is associative. Indeed, if \(f, g\), and \(h\) belong to \(\mathscr{L}^{1}(G)\), both functions \((f \times g) \times h\) and \(f \times(g \times h)\) belong to \(\mathscr{L}^{1}(G)\); and they are equal almost everywhere because the function $$ (x, y, z) \rightarrow f(z) g\left(z^{-1} y\right) h\left(y^{-1} z\right) $$ belongs to \(\mathscr{L}^{1}(G \times G \times G)\). The operation \(f \rightarrow f^{*}\) is con jugate linear and isometric with period two by 6.6.18. Moreover, we have $$ \begin{aligned} g^{*} \times f^{*}(x) &=\int g^{*}(y) f^{*}\left(y^{-1} x\right) d y \\ &=\int \overline{f\left(x^{-1} y\right)} \Delta\left(x^{-1} y\right) \overline{g\left(y^{-1}\right)} \Delta\left(y^{-1}\right) d y \\ &\left.=\int \overline{f(y)} \overline{g\left(y^{-1}\right.} x^{-1}\right) d y \Delta\left(x^{-1}\right)=(f \times g) *(x) \end{aligned} $$ which shows that \(*\) is antimultiplicative, and thus is an involution on \(L^{1}(G)\).

Even though the integration theory is at heart concerned only with real-valued functions, there is no difficulty in extending it to complex functions. We say that a function \(f: X \rightarrow \mathbb{C}\) is Borel/measurable provided that the same holds for the two functions \(\operatorname{Re} f\) and \(\operatorname{Im} f\). In fact, this is equivalent with the demand that \(f^{-1}(B) \in \mathscr{B}\) (respectively \(\mathscr{M}\) ) for every Borel subset \(B\) in \(\mathbb{C}\) (because it suffices to consider sets \(B\) of the form \(\\{\alpha \in \mathbb{C} \mid \operatorname{Re} \alpha>t\\}\) and \(\\{\alpha \in \mathbb{C} \mid \operatorname{Im} \alpha>t\\}, \mathrm{cf} .6 .2 .5\), and the counterimages of such sets depend only on \(\operatorname{Re} f\) or on \(\operatorname{Im} f)\). In the same manner we define the complex \(\mathscr{L}^{p}\)-spaces and the complex null functions as the complexifications of the corresponding real spaces. This means that the integral \(\int\) becomes a complex linear functional on the vector space of complex, integrable functions [which we still, with an absolute disregard for consistency, shall denote by \(\left.\mathscr{L}^{1}(X)\right]\). This extension process is painless, but the following complex version of \(6.1 .11\) and \(6.2 .16\) requires a special proof.

Theorem. If a function \(f\) on \(X\) is the pointwise limit of an increasing sequence \(\left(f_{n}\right)\) in \(\mathscr{L}^{1}(X)\) such that \(\sup \int f_{n}<\infty\), then \(f \in \mathscr{L}^{1}(X)\) and \(\int f=\lim \int f_{n}\).

Proposition. If \(f\left(f_{n}\right)\) is a sequence in \(\mathscr{L}^{1}(X)\) such that \(f_{n}(x) \leq f_{n+1}(x)\) for almost all \(x\) and every \(n\), and if \(\lim \int f_{n}<\infty\), there is an element \(f\) in \(\mathscr{L}^{1}(X)\) such that \(\int f=\lim \int f_{n}\) and \(f(x)=\lim f_{n}(x)\) almost everywhere. Proof. For each \(n\) there is a null set \(N_{n}\) such that \(f_{n}(x) \leq f_{n+1}(x)\) for \(x \notin N_{n}\). With \(N=\bigcup N_{n}\) we have a null set \(N\) and an extended-valued function \(f\) such that \(f_{n}(x) \nearrow f(x)\) for \(x \notin N\). Since6.4. \(L^{p}\)-Spaces 241 $$ \int\left[\left\\{f_{n}>m\right\\}\right] \leq \int m^{-1} f_{n} $$ for every \(n\) and \(m\), we see from \(6.1 .13\) that $$ \int[\\{f>m\\}] \leq m^{-1} \lim \int f_{n^{*}} $$ It follows that \(N_{\infty}=\\{f=\infty\\}\) is a null set, so that \(6.1 .13\) can be applied to the restriction of \(\left(f_{n}\right)\) and \(f\) to \(X \backslash\left(N_{\infty} \cup N\right)\).
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