Convergence in measure of the product of two functions

[SqueeSpleen]

$f_{k} \overset{m}{\rightarrow} f$ and $g_{k} \overset{m}{\rightarrow} g$ over $E$.
Then:
a)
$f_{k} + g_{k} \overset{m}{\rightarrow} f+g$ over $E$

b)
If $| E | < + \infty$, then $f_{k} g_{k} \overset{m}{\rightarrow} fg$ over $E$. Show that the hiphotesis $| E | < + \infty$ is neccesary

c) Let $\{ \frac{f_{k}}{g_{k}} \}_{k \in \mathbb{N}}$ a sequence of functions defined almost everywhere over $E$.
If $| E | \rightarrow g$ over $E$ and $g \neq 0$ almost everywhere, then $\frac{f_{k}}{g_{k}} \overset{m}{\rightarrow} \frac{f}{g}$ over $E$.

a)
$g_{k} \overset{m}{\rightarrow} g$ means
$\forall \delta > 0 , \forall \varepsilon > 0 , \exists k_{0} / \forall k \geq k_{0}:$
$| \{ | g - g_{k} | \geq \delta \} | < \varepsilon$
So for every $\frac{\delta}{2} > 0, \frac{\varepsilon}{2} > 0$ we have $k^{'}_{0}, k^{''}_{0}$ such that:

$\forall k \geq k^{'}_{0} : | \{ | g - g_{k} | \geq \frac{\delta}{2} \} | < \frac{\varepsilon}{2}$
$\forall k \geq k^{''}_{0} : | \{ | f - f_{k} | \geq \frac{\delta}{2} \} | < \frac{\varepsilon}{2}$
$k_{0} = max \{ k^{'}_{0}, k^{''}_{0} \}$
Now this holds for all $k \geq k_{0}$
Then
$\forall k \geq k_{0} : | \{ | g - g_{k} | \geq \frac{\delta}{2} \} \cup \{ | f - f_{k} | \geq \frac{\delta}{2} \} | < \varepsilon$
(The measure of the union is equal or lesser than sum of measures).
$\{ | (f - f_{k}) + (g - g_{k}) \geq \delta \} \subset \{ | f - f_{k} | + | g - g_{k} | \geq \delta \} \subset \{ | g - g_{k} | \geq \frac{\delta}{2} \} \cup \{ | f - f_{k} | \geq \frac{\delta}{2} \}$
Then for monotony of the measure:
$\forall k \geq k_{0} | \{ | (f - f_{k}) + (g - g_{k}) \geq \delta \} | < \varepsilon$

b) I guess I can prove:
$f_{k} \overset{m}{\rightarrow} f \Longrightarrow | f_{k} | \overset{m}{\rightarrow} |f |$
$f=f^{+}-f^{-}$ and $| f | = f^{+} + f^{-}$ (so it's easy, we only need to prove that if $f_{k} \overset{m}{\rightarrow} f$ then $f^{\mp}_{k} \overset{m}{\rightarrow} f^{\mp}$
Then using that prove that if $f_{k} \overset{m}{\rightarrow} f$ then $f^{2}_{k} \overset{m}{\rightarrow} f^{2}$ (*)
And finally notice that $fg = \frac{1}{4} [ (f+g)^{2} - (f-g)^{2} ]$ (Using also that if $f_{k} \overset{m}{\rightarrow} f$ then $-f_{k} \overset{m}{\rightarrow} -f$)

(*)
$| f_{k} | \overset{m}{\rightarrow} | f |$ means
$\forall \delta > 0 , \forall \varepsilon > 0 , \exists k_{0} / \forall k \geq k_{0}:$
$| \{ | | f | - | f_{k} | | \geq \delta \} | < \varepsilon$

What I thought was:
$f^{2} - f^{2}_{k} = (\underbrace{f+f_{k}}_{\text{if }f <+\infty \text{ it tends to }2f})(\underbrace{f-f_{k}}_{ \to 0})$
And later do something like this:
$| \{ | f - f_{k} | \geq \frac{\delta}{f+f_{k}} \} | < \varepsilon$
But I realized that this doesn't make any sense at all because $f+f_{k}$ depends of the value of $x$.

Any hint?

 

[lippyka]

c)g_{k} \overset{m}{\rightarrow} g means\forall \delta > 0 , \forall \varepsilon > 0 , \exists k_{0} / \forall k \geq k_{0}: | \{ | g - g_{k} | \geq \delta \} | < \varepsilon So for every \frac{\delta}{2} > 0, \frac{\varepsilon}{2} > 0 we have k^{'}_{0}, k^{''}_{0} such that:\forall k \geq k^{'}_{0} : | \{ | g - g_{k} | \geq \frac{\delta}{2} \} | < \frac{\varepsilon}{2} \forall k \geq k^{''}_{0} : | \{ | f - f_{k} | \geq \frac{\delta}{2} \} | < \frac{\varepsilon}{2} k_{0} = max \{ k^{'}_{0}, k^{''}_{0} \}Now this holds for all k \geq k_{0}Then\forall k \geq k_{0} : | \{ | g - g_{k} | \geq \frac{\delta}{2} \} \cup \{ | f - f_{k} | \geq \frac{\delta}{2} \} | < \varepsilon (The measure of the union is equal or lesser than sum of measures).\{ | \frac{f - f_{k}}{g - g_{k}} \geq \delta \} \subset \{ | f - f_{k} | + | g - g_{k} | \geq \delta \} \subset \{ | g - g_{k} | \geq \frac{\delta}{2} \} \cup \{ | f - f_{k} | \geq \frac{\delta}{2} \}Then for monotony of the measure:\forall k \geq k_{0}