Chapter 2

If \(f_{n} \rightarrow f\) almost uniformly, then \(f_{n} \rightarrow f\) a.e. and in measure.

Let \(\mu\) be counting measure on \(\mathrm{N}\). Then \(f_{n} \rightarrow f\) in measure iff \(f_{n} \rightarrow f\) uniformly.

Let \(X=Y\) be an uncountable linearly ordered set such that for each \(x \in X\), \(\\{y \in X: y

Let \(X=Y=N, \mathcal{M}=\mathcal{N}=\mathcal{P}(\mathrm{N}), \mu=\nu=\) counting measure. Define \(f(m, n)=1\) if \(m=n, f(m, n)=-1\) if \(m=n+1\), and \(f(m, n)=0\) otherwise. Then \(\int|f| d(\mu \times \nu)=\infty\), and \(\iint f d \mu d \nu\) and \(\iint f d \nu d \mu\) exist and are unequal.

Suppose \((X, \mathcal{M}, \mu)\) is a \(\sigma\)-finite measure space and \(f \in L^{+}(X) .\) Let $$ G_{f}=\\{(x, y) \in X \times[0, \infty]: y \leq f(x)\\} $$ Then \(G_{f}\) is \(\mathrm{M} \times \mathrm{B}_{\mathrm{R}}\)-measurable and \(\mu \times m\left(G_{f}\right)=\int f d \mu ;\) the same is also true if the inequality \(y \leq f(x)\) in the definition of \(G_{f}\) is replaced by \(y

Let \((X, \mathcal{M}, \mu)\) and \((Y, \mathcal{N}, \nu)\) be arbitrary measure spaces (not necessarily \(\sigma\) finite). a. If \(f: X \rightarrow \mathbb{C}\) is \(\mathcal{M}\)-measurable, \(g: Y \rightarrow \mathbb{C}\) is N-measurable, and \(h(x, y)=\) \(f(x) g(y)\), then \(h\) is \(\mathcal{M} \otimes \mathcal{N}\)-measurable. b. If \(f \in L^{1}(\mu)\) and \(g \in L^{1}(\nu)\), then \(h \in L^{1}(\mu \times \nu)\) and \(\int h d(\mu \times \nu)=\) \(\left[\int f d \mu\right]\left[\int g d \nu\right]\).

Let \(E=[0,1] \times[0,1]\). Investigate the existence and equality of \(\int_{E} f d m^{2}\), \(\int_{0}^{1} \int_{0}^{1} f(x, y) d x d y\), and \(\int_{0}^{1} \int_{0}^{1} f(x, y) d y d x\) for the following \(f\). a. \(f(x, y)=\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}\right)^{-2}\). b. \(f(x, y)=(1-x y)^{-a}(a>0)\). c. \(f(x, y)=\left(x-\frac{1}{2}\right)^{-3}\) if \(0

If \(f\) is Lebesgue integrable on \((0, a)\) and \(g(x)=\int_{x}^{a} t^{-1} f(t) d t\), then \(g\) is integrable on \((0, a)\) and \(\int_{0}^{a} g(x) d x=\int_{0}^{a} f(x) d x\).

Show that \(\int e^{-s x} x^{-1} \sin ^{2} x d x=\frac{1}{4} \log \left(1+4 s^{-2}\right)\) for \(s>0\) by integrating \(e^{-s x} \sin 2 x y\) with respect to \(x\) and \(y\).

Let \(f(x)=x^{-1} \sin x\). a. Show that \(\int_{0}^{\infty}|f(x)| d x=\infty\). b. Show that \(\lim _{b \rightarrow \infty} \int_{0}^{b} f(x) d x=\frac{1}{2} \pi\) by integrating \(e^{-x y} \sin x\) with respect to \(x\) and \(y\). (In view of part (a), some care is needed in passing to the limit as \(b \rightarrow \infty .)\)