# Thread: measurable set on R^2

1. ## 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.

2. Originally Posted by Archi
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 $\phi\psi = \tfrac12\bigl((\phi + \psi)^2 - \phi^2 - \psi^2\bigr)$.

3. Originally Posted by Opalg
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?

4. Originally Posted by Archi
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 $(x,y)\mapsto f(x-y)g(y)$.

5. Originally Posted by Opalg
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?

6. Originally Posted by Archi
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.)

Originally Posted by Archi
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.