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Math Help - measurable set on R^2

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    measurable set on R^2

    Let f,g:R\rightarrow R be Lebesgue integrable functions on the real line R and let h(x)=\int_R f(x-y)g(y)dy, x \in R.
    Show that f(x-y)g(y) is measurable on R.

    i dont even know how to start. help is appreciated so much.
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    Quote Originally Posted by Archi View Post
    Let f,g:R\rightarrow R be Lebesgue integrable functions on the real line R and let h(x)=\int_R f(x-y)g(y)dy, x \in R.
    Show that f(x-y)g(y) is measurable on R.
    If f and g are integrable then (by definition) they must be measurable.

    Start by showing that the function y\mapsto f(x-y) is measurable (for fixed x). You should be able to do that from the definition of measurability. You then need to show that the product of that function with g is measurable. To show that the product of two measurable functions is measurable, first show (from the definition) that if \phi is measurable then so is \phi^2. Then, if \phi and \psi are measurable, so is \phi\psi = \tfrac12\bigl((\phi + \psi)^2 - \phi^2 - \psi^2\bigr).
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    Quote Originally Posted by Opalg View Post
    If f and g are integrable then (by definition) they must be measurable.

    Start by showing that the function y\mapsto f(x-y) is measurable (for fixed x). You should be able to do that from the definition of measurability. You then need to show that the product of that function with g is measurable. To show that the product of two measurable functions is measurable, first show (from the definition) that if \phi is measurable then so is \phi^2. Then, if \phi and \psi are measurable, so is \phi\psi = \tfrac12\bigl((\phi + \psi)^2 - \phi^2 - \psi^2\bigr).
    Sorry what i wanted to ask was to show that f(x-y)g(y) is measurable on R^2. so x and y are both variables right? would u help me on this, please?
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    Quote Originally Posted by Archi View Post
    Sorry what i wanted to ask was to show that f(x-y)g(y) is measurable on R^2. so x and y are both variables right? would u help me on this, please?
    If E is a measurable subset of \mathbb{R}, then \{(x,y):x-y\in E\} is a measurable subset of \mathbb{R}^2. Reason: it is equal to the set \mathbb{R}\times (E/\sqrt2) (which is measurable) rotated through an angle \pi/4, and rotations preserve measurability. Therefore if f is measurable, so is the function (x,y)\mapsto f(x-y). The function (x,y)\mapsto g(y) is clearly measurable. Hence so is the product (x,y)\mapsto f(x-y)g(y).
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    Quote Originally Posted by Opalg View Post
    If E is a measurable subset of \mathbb{R}, then \{(x,y):x-y\in E\} is a measurable subset of \mathbb{R}^2. Reason: it is equal to the set \mathbb{R}\times (E/\sqrt2) (which is measurable) rotated through an angle \pi/4, and rotations preserve measurability. Therefore if f is measurable, so is the function (x,y)\mapsto f(x-y). The function (x,y)\mapsto g(y) is clearly measurable. Hence so is the product (x,y)\mapsto f(x-y)g(y).
    i cant see the 45 degree rotation by multiplying E by 1/\sqrt2. but in this question, \{(x,y):x-y\in E\}=\{(x,y):x-y \in f^{-1}(\alpha,\infty)\} for any \alpha, is it correct? then \{(x,y):x-y \in f^{-1}(\alpha,\infty)\}=\{(x,y):x-y>\alpha\}=R\times (-\infty,x-\alpha). since R and (-\infty,x-\alpha) are both measurable for each \alpha. so R\times (-\infty,x-\alpha) is measurable. can u tell me if what i am doing is correct, please?
    Last edited by Archi; December 29th 2009 at 01:14 PM. Reason: wrong notation
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  6. #6
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    Quote Originally Posted by Archi View Post
    i cant see the 45 degree rotation by multiplying E by 1/\sqrt2.
    The reason for the 45 degrees is that the function (x,y)\mapsto f(x-y) is obviously constant along lines of the form x-y = const. So if you rotate the set \{(x,y):f(x-y)>\alpha\} through 45 degrees then you get a set that is constant along lines parallel to one of the axes. (The factor 1/\sqrt2 comes in because it is the cosine of 45 degrees.)

    Quote Originally Posted by Archi View Post
    but in this question, \{(x,y):x-y\in E\}=\{(x,y):x-y \in f^{-1}(\alpha,\infty)\} for any \alpha, is it correct? Yes, correct so far.

    then \{(x,y):x-y \in f^{-1}(\alpha,\infty)\}=\{(x,y):x-y>\alpha\} No, that is not correct. It is correct that \color{red}\{(x,y):x-y \in f^{-1}(\alpha,\infty)\}=\{(x,y):f(x-y)>\alpha\}, but you can't just make the f disappear.
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