Ex4-1

Poblem: Prove the following:
(i) Corollary 4.2: If f is measurable, then\{f>-\infty\}, \{f<+\infty\}, \{f=+\infty\}, \{a\leq f\leq b\}, \{f=a\} etc. are all measurable, Moreover, f is measurable if and only if \{a<f<+\infty\} is measurable for every finite a.
(ii) Theorem 4.8: If f is measurable and \lambda is any real number, then f+\lambda, \text{and} \lambda f are measurable.
Proof: (i)
First note that \{f>-\infty\}=\cup_{k=1}^{\infty}\{f>-k\} and \{f<+\infty\}=\cup_{k=1}^{\infty}\{f<k\}, both of these sets are measurable each of them is the union of the countable measurable set. Now, note that the set \{f=+\infty\}=\{f<+\infty\}^{c} is measurable as the complement of a measurable set. After that, we also have
\{a\leq f\leq b\}=(\cap_{k=1}^{\infty}\{f>a-\frac{1}{k}\})\cap\{f>b\}^{c}
this shows that the set is measurable since it is intersection of countable union of measurable set. Lastly, we have \{f=a\}=\{a\leq f\leq a\} is measurable.
For the second statement, the forward implication is implied by the above, since
\{a<f<+\infty\}=\{f>a\}\cap\{f<+\infty\}
is measurable . Now, for another direction, assume that all the sets \{a<f<+\infty\} are measurable: then
\cup_{k=1}^{\infty}\{-k<f<+\infty\}=\{f<+\infty\}
is measurable, so 
\{f=+\infty\}=\{f<+\infty\}^c
is measurable, so
\{f>a\}=\{a<f<+\infty\}\cup\{f=+\infty\}
is measurable. Thus, f is measurable.
(ii) If f is measurable, then \{f>a-\lambda\}=\{f+\lambda>a\} is a measurable set for all a. If \lambda>0, then \{f>a/\lambda\}=\{\lambda f>a\} is measurable for all a. If \lambda<0, then \{f<a/\lambda\}=\{\lambda f>a\} is measurable for all a. If \lambda=0, then \lambda f\equiv0 identically, which is measurable. Thus f+\lambda and \lambda f are measurable function for all \lambda\in\mathbb{R}.