Convergence in measure of the product of two functions

In summary, the conversation discusses the concept of convergence in measure and its application in proving the convergence of functions over a given set. It also introduces the necessary hypothesis of the set having finite measure.
  • #1
SqueeSpleen
141
5
[itex]f_{k} \overset{m}{\rightarrow} f[/itex] and [itex]g_{k} \overset{m}{\rightarrow} g[/itex] over [itex]E[/itex].
Then:
a)
[itex]f_{k} + g_{k} \overset{m}{\rightarrow} f+g[/itex] over [itex]E[/itex]

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

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

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

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

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

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

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

Any hint?
 
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  • #2
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}
 

FAQ: Convergence in measure of the product of two functions

What does convergence in measure of the product of two functions mean?

Convergence in measure of the product of two functions refers to the concept in measure theory where the product of two measurable functions converges to a limit in measure. This means that the measure of the set where the product of the functions differ from the limit approaches zero as the measure of the set itself approaches zero.

What is the significance of convergence in measure of the product of two functions?

Convergence in measure of the product of two functions is an important concept in measure theory as it allows us to prove convergence of products of functions under certain conditions. It also has applications in various areas of mathematics, such as probability theory and functional analysis.

What are some examples of functions that exhibit convergence in measure?

Some common examples of functions that exhibit convergence in measure are bounded functions, uniformly integrable functions, and Lp integrable functions. These functions satisfy certain conditions that guarantee convergence in measure of their product.

What is the relationship between convergence in measure and other types of convergence?

Convergence in measure is a weaker form of convergence compared to other types of convergence, such as pointwise convergence or almost everywhere convergence. This means that if a sequence of functions converges in measure, it may not necessarily converge pointwise or almost everywhere.

How is convergence in measure of the product of two functions related to the product of their limits?

If two sequences of functions converge in measure to their respective limits, then the product of these functions will also converge in measure to the product of their limits. However, the converse is not always true, as the product of two functions can converge in measure even if the individual functions do not converge in measure.

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