Chapter 2: Problem 39
If \(f_{n} \rightarrow f\) almost uniformly, then \(f_{n} \rightarrow f\) a.e. and in measure.
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Suppose that \(f_{n}\) and \(f\) are measurable complex-valued functions and \(\phi: \mathbb{C} \rightarrow C\). a. If \(\phi\) is continuous and \(f_{n} \rightarrow f\) a.e., then \(\phi \circ f_{n} \rightarrow \phi \circ f\) a.e. b. If \(\phi\) is uniformly continuous and \(f_{n} \rightarrow f\) uniformly, almost uniformly, or in measure, then \(\phi \circ f_{n} \rightarrow \phi \circ f\) uniformly, almost uniformly, or in measure, respectively. c. There are counterexamples when the continuity assumptions on \(\phi\) are not satisfied.
Suppose that \(f\) is a function on \(\mathbb{R} \times \mathbb{R}^{k}\) such that \(f(x, \cdot)\) is Borel measurable for each \(x \in \mathbb{R}\) and \(f(\cdot, y)\) is continuous for each \(y \in \mathbb{R}^{k}\). For \(n \in \mathbb{N}\), define \(f_{n}\) as follows. For \(i \in \mathbb{Z}\) let \(a_{i}=i / n\), and for \(a_{i} \leq x \leq a_{i+1}\) let $$ f_{n}(x, y)=\frac{f\left(a_{i+1}, y\right)\left(x-a_{i}\right)-f\left(a_{i}, y\right)\left(x-a_{i+1}\right)}{a_{i+1}-a_{i}} $$ Then \(f_{n}\) is Borel measurable on \(\mathbb{R} \times \mathbb{R}^{k}\) and \(f_{n} \rightarrow f\) pointwise; hence \(f\) is Borel measurable on \(\mathbb{R} \times \mathbb{R}^{k}\). Conclude by induction that every function on \(\mathbb{R}^{n}\) that is continuous in each variable separately is Borel measurable.
Let \(\mu\) be counting measure on \(\mathbb{N}\). Interpret Fatou's lemma and the monotone and dominated convergence theorems as statements about infinite series.
Suppose \(\mu(X)<\infty\). If \(f\) and \(g\) are complex-valued measurable functions on \(X\), define $$ \rho(f, g)=\int \frac{|f-g|}{1+|f-g|} d \mu $$ Then \(\rho\) is a metric on the space of measurable functions if we identify functions that are equal a.e., and \(f_{n} \rightarrow f\) with respect to this metric iff \(f_{n} \rightarrow f\) in measure.
If \(f \in L^{+}\), let \(\lambda(E)=\int_{E} f d \mu\) for \(E \in \mathcal{M}\). Then \(\lambda\) is a measure on \(\mathcal{M}\), and for any \(g \in L^{+}, \int g d \lambda=\int f g d \mu\). (First suppose that \(g\) is simple.)
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