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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

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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

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)

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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. – Evgeny Savinov 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) – Evgeny Savinov Oct 19 '11 at 16:00
The fact that you can cover by this method follows from bijectable and continous function $f(x)$ – Evgeny Savinov Oct 19 '11 at 16:11

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