5-2

Question:
(a) Show that the conclusions of Theorem 5.32 are not generally true without the assumption that  \phi \in L(E).
(b) Show that Theorem 5.33 fails without the assumption that |E| < \infty.
Solution:
(a) For (i), Consider the function f_{k=}\left\{\begin{matrix} 
 -\infty \ ,x>k\\ 
 0 \ \ ,x\le k
\end{matrix}\right. and \lim_{k \to \infty} f_{k}= 0 and take E=R.
In this case, \int _{E} \lim_{k \to \infty} f_{k}= \int _{E} 0=0.
Since \int _{E} f_{k} = -\infty for all k, we have \lim_{k \to \infty} \int _{E} f_{k}= -\infty . You could see that \lim_{k \to \infty} \int _{E} f_{k} \ne \int _{E} \lim_{k \to \infty} f_{k} .
For (ii), i think the function -f_{k=}\left\{\begin{matrix} 
 \infty \ ,x>k\\ 
 0 \ \ ,x\le k
\end{matrix}\right. is a good counter—example, and this proof is similar to the proof above.
(b) Consider the density function of exponential distribution which comes from probability. It's f_{k=}\left\{\begin{matrix} 
 \frac{1}{k} e^{-\frac{1}{k}x} \ ,x>0\\ 
 0 \ \ ,x\le 0
\end{matrix}\right. and \lim_{k \to \infty} f_{k} =0 . And we take E=R .
We claim that f_{k} uniformly converges to f(x)=0 first.
We take N=\lfloor\frac{1}{ \varepsilon }\rfloor+1 , then for all x \in R , |f_{k}-f| \le \frac{1}{k} e^{-\frac{1}{k}x} \le \frac{1}{k} e^{-\frac{1}{k} \times 0} = \frac{1}{k} \le \frac{1}{\lfloor\frac{1}{ \varepsilon }\rfloor+1} \le \frac{1}{\frac{1}{\varepsilon}}=\varepsilon for all k \ge N , which shows that f_{k} uniformly converges to f .
In this case, \int _{E} f_{k} = 1 for all k and therefore \lim_{k \to \infty} \int _{E} f_{k} = 1 .
But \int _{E} f = \int _{E} 0 =0 , and you could see that \lim_{k \to \infty} \int _{E} f_{k} \ne \int _{E} f in this case.
Note:
Theorem 5.32: (Monotone Convergence Theorem) Let \left \{ f_{k} \right \} be a sequence of measurable functions on E:
(i) If f_{k} \nearrow f a.e. on E and there exists φ ∈ L(E) such that f_{k} ≥ φ a.e. on E for all k, then \int_{E} f_{k} \rightarrow \int_{E} f .
(ii)If f_{k} \searrow f a.e. on E and there exists φ ∈ L(E) such that f_{k} ≤ φ a.e. on E for all k, then \int_{E} f_{k} \rightarrow \int_{E} f .
Theorem  5.33: (Uniform Convergence Theorem) Let f_{k} \in L(E) for k=1,2, \ldots , and let \left\{f_{k}\right\} converge uniformly to f on E, |E|<+\infty . Then f \in L(E) and \int_{E} f_{k} \rightarrow \int_{E} f .