I want to show that $(f*g)(x)=\int_{\mathbb{R}^n}f(t)g(t-x)dt$ is well defined where $f\in L^1(\mathbb{R}^n)$, $g\in L^\infty(\mathbb{R}^n)$.

that is, I need to show that $f(t)g(t-x)$ is integrable in $t$ for a.e. $x$.

and here, I was seggested to use the Fubini's/ or Tonelli's theorem, but I don't know exactly how to do it.

by the wiki, Fubini's theorem is

if $X,Y$ are $\sigma$-finite measure spaces, and if $f$ is measurable and $\int_{X\times Y} |f(x,y)|d(x,y)<\infty$, then $\int_{X\times Y} f(x,y)d(x,y)=\int_X\int_Yf(x,y)dydx=\int_Y\int_Xf(x,y)dxdy$.

since $\mathbb{R}=\cup_{t=0}^\infty[(-1-t,-t]\cup[t,1+t)]$, and $\mathbb{R}^n=\mathbb{R}\times...\times \mathbb{R}$ ($n$ times), $\mathbb{R}^n$ is $\sigma$-finite(this condition is needed to make the product measure unique but I don't know how exactly).

now let $h(x,t)=f(t)g(t-x)$. and I don't even know how to show that $h$ is measurable in $X\times Y=\mathbb{R}^n\times \mathbb{R}^n$ and the integral of $|h|$ is finite.

so, instead of this, I tried to use the Fubini-Tonelli's theorem, which is

if $X,Y$ are $\sigma$-finite measure spaces, and if $f$ is measuarable such that at least one of $\int_{X\times Y} |f(x,y)|d(x,y)$, $\int_X\int_Y|f(x,y)|dydx$, $\int_Y\int_X|f(x,y)|dxdy$ is finite, then $\int_{X\times Y} f(x,y)d(x,y)=\int_X\int_Yf(x,y)dydx=\int_Y\int_Xf(x,y)dxdy$.

then, $\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|h(x,t)|dtdx=\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|f(t)g(t-x)|dtdx \leq||g||_\infty\int_{\mathbb{R}^n}\int_{\mathbb{R}^n}|f(t)|dtdx=||g||_\infty\int_{\mathbb{R}^n}||f||_1dx=||g||_\infty||f||_1\int_{\mathbb{R}^n}dx=+\infty$

and I got the same result with $dx,dt$ exchanged. so the only option that I have is to show that $h$ is measurable in $\mathbb{R}^n\times\mathbb{R}^n$ and $\int_{\mathbb{R}^n\times\mathbb{R}^n}|h(x,t)|d(x,t)<\infty$, the same assumption with the original Fubini.

anyway, let say the Fubini's theorem is applicable. then $h(x,t)$ is integrable in $t$ for a.e. $x$. is this the correct result? then how'm I supposed to satisfy the assumption of Fubini?

  • $\begingroup$ This is a direct application of Hölder's inequality. $\endgroup$ – Michael Lee Jun 20 '17 at 14:35
  • $\begingroup$ @MichaelLee How is it so? can you explain? $\endgroup$ – user159234 Jun 20 '17 at 14:39

I recommend instead that you use Hölder's inequality:

$$\left\lvert \int_{\mathbb{R}^n} f(t)g(t-x)\,dt\right\rvert\leq \int_{\mathbb{R}^n} \lvert f(t)g(t-x)\rvert\,dt\leq \|g\|_{L^{\infty}(\mathbb{R}^n)}\int_{\mathbb{R}^n} \lvert f(t)\rvert\,dt < \infty$$

Therefore, $f(t)g(t-x)$ is integrable in $t$. Hölder's inequality follows from either Young's inequality or Jensen's inequality (you can see the linked page for a proof).

  • $\begingroup$ don't we need to verify first that whether $f(t)g(t-x)$ is measurable in $t$ or not? because integrable means measurable and finite integral? $\endgroup$ – user159234 Jun 20 '17 at 14:47
  • $\begingroup$ The product of measurable functions is measurable, of course. Both $f$ and $g$ are measurable by the fact that they reside in $L^1$ and $L^{\infty}$, respectively, and $g$ is still measurable when we translate its argument by the preimage definition of measurability. $\endgroup$ – Michael Lee Jun 20 '17 at 14:49
  • $\begingroup$ that is exactly what I don't know. in this case we have two variables and everything is mixed up. how can I show the product is measurable? $\endgroup$ – user159234 Jun 20 '17 at 14:51
  • $\begingroup$ I edited my comment to include the details. $\endgroup$ – Michael Lee Jun 20 '17 at 14:53
  • $\begingroup$ ah we fix $x$. thank you. it's all clear now. $\endgroup$ – user159234 Jun 20 '17 at 14:54

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