Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Here are two

i) Let $ g:[c,d] \to R $ be continuous and $ f:[a,b] \to [c,d] $ integrable , then $ g\left( {f\left( x \right)} \right):[a,b] \to R $ it´s also integrable.

ii) $ f:[a,b] \to [c,d] $, $ f \in C^1 $, $ f´\left( x \right) \ne 0 $ for every x $ \in [a,b] $ and $ g:[c,d] \to R $ integrable, then again $ g\left( {f\left( x \right)} \right) $ it´s integrable on [a,b]

How can I do this problem, I suppose that the characterization under measure of the discontinuity set, will help, but I don´t know how to use it <.< , sorry for ask this basic things, but I´m starting to learn

share|improve this question
    
In the second problem, the conditions imply that $f$ is strictly monotone, hence invertible and the inverse is strictly monotone and $\mathcal{C}^1$. Do you have any results about whether how such functions (perhaps on finite closed intervals) work on sets of measure $0$? If you can prove that the inverse image under $f$ of the points of discontinuity of $g$ is a null set you'll get the result you want. –  Arturo Magidin Oct 19 '11 at 4:58

1 Answer 1

The first: If $f(x)$ is continuous in $x_0$, then $g(f(x))$ is also continuous in this point as $g(x)$ is continuous (Theorem about composite function). It leads to statement: points of discontinuity $g(f(x))$ can be only points of discontinuity $f(x)$. But as $f(x)$ is integrable, it has Lebesgue measure of its points of discontinuity (Lebesgue integrability condition) equal to $0$ $\Rightarrow$ $g(f(x))$ has Lebesgue measure of its points of discontinuity equal to $0$. It means that $g(f(x))$ is integrable (Lebesgue integrability condition). QED.

The second: As $g(x)$ is integrable then its Lebesgue measure of its points of discontinuity equal $0$, as $f(x)$ is monotonic ($f'(x)\not=0$) then $f^{-1}$ exists on $[c,d]$ and its also monotonic (Theorem about inverse function). It means that each point of discontinuity has single inverse image point on $[a,b]$. If the set of points of discontinuity $g(x)$ has measure zero, for any $\epsilon>0$ we can cover this set by countable number of intervals of total length $(tl_1)$$<\epsilon$. Then let's get center of some interval ($I_k$) and find its inverse image, and cover this point $(x_0)$ by the interval of length $=2\frac{\text{length}(I_1)}{|f'(x_0)|}$, then do it with every interval, so now we have covering of set of discontinuity point $g(f(x))$ by countable number of intervals, let's check its total length $(tl_2)$,$tl_2<=2*tl_1/\min[|f'(x)|]=C*tl_1$; $(f'(x)$ is continuous and $f'(x)\not=0$ => for any $x$ from segment $|f'(x)|>r>0$). It leads to $g(f(x))$ Lebesgue measure of its points of discontinuity equal to $0$ $\Rightarrow$ $g(f(x))$ is integrable (Lebesgue integrability condition)

share|improve this answer
    
Not every set that is bijectable with a set of measure zero is of measure zero (e.g., the cantor set is bijectable with the entire unit interval), so there is a bit of work to be done beyond "each point of discontinuity has a single inverse image point on $[a,b]$", I think... –  Arturo Magidin Oct 19 '11 at 5:09
    
Yeh, I understood. It'll be fixed. –  Savinov Evgeny Oct 19 '11 at 6:47
    
@SavinovEvgeny Thanks!, but I have a question about the second problem, How can I prove that the second really cover , all the points of discontinuity, and Why you choosed this length to do it. I don´t understand nothing )= –  Susuk Oct 19 '11 at 14:21
    
if we get some segment [p,q] which length is sufficiently small $f:[p,q]->[f(p),f(q)]$. Using the theorem of Lagrange $f(p)-f(q)=f'(r)*(p-q)$ and $r$ belongs to $(p,q)$, then $\frac{|p-q|}{|f(p)-f(q)|}=\frac{1}{|f'(r)|}<2\frac{1}{|f'(\frac{p+q}{2})|}$ (We can say it because $f'(x)$ is continuous) –  Savinov Evgeny Oct 19 '11 at 16:00
    
The fact that you can cover by this method follows from bijectable and continous function $f(x)$ –  Savinov Evgeny Oct 19 '11 at 16:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.