Ex4-17

Question: Suppose that f_k\xrightarrow{m}f and g_k\xrightarrow{m}g on E. Show that f_k+g_k\xrightarrow{m}f+g on E and, if |E|<+\infty, that f_kg_k\xrightarrow{m}fg on E. If, in addition, g_k\rightarrow g on E, g\neq0 a.e., and |E|<+\infty, show that f_k/g_k\xrightarrow{m}f/g on E. (For the product f_kg_k, write f_kg_k-fg=(f_k-f)(g_k-g)+f(g_k-g)+g(f_k-f). Consider each term separately, using the fact that a function that is finite on E, |E|<+\infty is bounded outside a subset of E with small measure.)
Answer:
Addition: 
Assume that f_k\xrightarrow{m}f and g_k\xrightarrow{m}g on E. Let \epsilon>0. There exists K_1 such that |\{|f-f_k|>\epsilon/2\}|<\epsilon/2 if k>K_1. Similarly, there exists K_2 such that |\{|g-g_k|>\epsilon/2\}|<\epsilon/2 if k>K_2.
Since \{|(f+g)-(f_k+g_k)|>\epsilon\}\subseteq\{|f-f_k|>\epsilon/2\}\cup\{|g-g_k|<\epsilon/2\}, thus |\{|(f+g)-(f_k+g_k)|>\epsilon\}|<\epsilon/2+\epsilon/2=\epsilon if k>\max\{K_1,K_2\}.
This implies f_k+g_k\xrightarrow{m}f+g on E.
Multiplication:
Further assume |E|<\infty. Write f_kg_k-fg=(f_k-f)(g_k-g)+f(g_k-g)+g(f_k-f).
There exists K_3 such that |\{|f-f_k|>\sqrt{\epsilon}\}|<\epsilon/2 if k>K_3. There exists K_4 such that |\{|g-g_k|>\sqrt{\epsilon}\}|<\epsilon/2 if k>K_4.
If k>\max\{K_3,K_4\}, then |\{|(f_k-f)(g_k-g)|>\epsilon\}|\leq|\{|f-f_k|>\sqrt{\epsilon}\}|+|\{|g-g_k|>\sqrt{\epsilon}\}|<\epsilon.
Thus (f_k-f)(g_k-g)\xrightarrow{m}0.
Applying Question 15 to \{f\}, there is a closed F\subseteq E and a finite M such that |E\backslash F|<\epsilon/2 and |f(x)|\leq M for all x\in F. We may assume M\neq 0, as the case M=0 is trivially true since |\{|f(g_k-g)|>\epsilon\}|=|\{x\in F:|f(g_k-g)|>\epsilon\}|+|\{x\in E\backslash F:|f(g_k-g)|>\epsilon\}|<0+\epsilon/2.
Thus |\{|f(g_k-g)|>\epsilon\}|=|\{x\in F:|f(g_k-g)|>\epsilon\}|+|\{x\in E\backslash F:|f(g_k-g)|>\epsilon\}|
\leq|\{x\in F:|g_k-g|>\epsilon/M\}|+|E\backslash F|<
\epsilon/2+\epsilon/2 for sufficiently large k. Thus f(g_k-g)\xrightarrow{m}0. Similarly g(f_k-f)\xrightarrow{m}0.
By our result of "Addition", f_kg_k-fg\xrightarrow{m}0, that is, f_kg_k\xrightarrow{m}fg on E.
Division:
Assume in addition that g_k\rightarrow g on E, g\neq0 a.e., and |E|<\infty. Note that since we have proved "Multiplication", it suffices to show that 1/g_k\xrightarrow{m}1/g on E.
We use Theorem 4.21 (a.e. convergence and |E|<\infty implies convergence in measure). Note that since g\neq0 a.e., 1/g is measurable and finite a.e. in E. Since g_k\rightarrow g on E, for sufficiently large k, g_k\neq0 a.e. so that 1/g_k is also measurable and finite a.e. in E.
By Theorem 4.21, since 1/g_k\rightarrow1/g a.e. on E and |E|<\infty, then 1/g_k\xrightarrow{m}1/g on E.