0

For a finite measure subset $A$, and two functions $h(x)$, $g(x)$ on $A$,

if $h(x)-g(y) \in L^1(A \times A)$, show that $h(x), g(x) \in L^1(A)$

ISO
  • 225
  • The question makes no sense. How can you have $h,g$ functions on $A$ but $h+g$ defined on $A\times A$? Unless you mean the function $f(x,y)=h(x)+g(y)$ or something to that effect. Anyway, how are you going from $\int_A|h(x)+g(x)|,dx=\int_{A\times A}|h(x)+g(x)|,dx,dy$? And how did you get the last $<\infty$? You’re already assuming $g\in L^1(A)$. Assuming you really mean the function $f$, I would argue by Fubini-Tonelli. – peek-a-boo Mar 19 '23 at 07:02
  • @peek-a-boo fixed. – ISO Mar 19 '23 at 07:15
  • 1
    There is no issue with $1/\sqrt{x}$ near the origin. Compute the integrals and check all three are finite. To prove it in general, note that by Tonelli, $\int_{A\times B}|f(x,y)|,d(x,y)=\int_A\int_B|f(x,y)|,dy,dx=\int_B\int_A|f(x,y)|,dx,dy$. So if the first is finite, then all three are finite. Now, what can you say about the inner integrals? – peek-a-boo Mar 19 '23 at 07:16
  • If that was too cryptic, then consider this: if $\int_X|\phi(x)|,dx<\infty$, then can $\phi$ be infinite at some points? (how many at worst)? Therefore… – peek-a-boo Mar 19 '23 at 07:20
  • @peek-a-boo is it a fact that if $\int_{A \times B} |f(x,y|d(x,y) < \infty$ then WLOG $\int_{A} |f(x,y|dx$ is also finite? – ISO Mar 19 '23 at 07:31
  • there are two hints I gave you to determine whether or not that is a true statement. – peek-a-boo Mar 19 '23 at 07:31
  • @peek-a-boo so it is true a.e.? Which theorem is that? – ISO Mar 19 '23 at 07:33
  • true for a.e $y\in B$, yes. It is the obvious statement that if $\phi\in L^1(X,\mu)$, then $E={x\in X,:, |\phi(x)|=\infty}$ has measure zero. The contrapositive may sound even more obvious: if that set has positive measure, then by monotonicity of integrals, $\int_X|\phi|,d\mu\geq\int_E|\phi|,d\mu=\infty\cdot\mu(E)=\infty$. – peek-a-boo Mar 19 '23 at 07:35
  • @peek-a-boo. got it. Another way is this Let $E={x \in X, |\phi(x)|=\infty}$. Note that $|\phi| \chi_{E}\leq |\phi|$, hence $\infty \cdot\mu(E)\leq \int |\phi| < \infty$

    That implies $\mu(E)=0$

     – ISO Mar 19 '23 at 07:36

0 Answers0