3
$\begingroup$

Let $(\Omega,\mathcal{F},\mu)$ be a measure space and $f,g:\Omega\to[0,\infty]$ be two non-negative measurable functions. If $\forall A\in\mathcal{F}: \int_A fd\mu\geq \int_A gd\mu$, can we conclude $f\geq g \, \mu-$a.e.?

I know that in case $f,g:\Omega\to\mathbb{R}$ and $f,g\in L^1(\mu)$ the result holds, since $\infty-\infty$ does not happen and the theorem becomes equivalent to $\forall A\in\mathcal{F}: \int_A f-gd\mu\geq0\implies f-g\geq0\, \mu-$a.e., which I know the proof.

Is this still true without $L^1$ assumption for non-negative measurable functions, where integral can become $+\infty$ and how can I prove that?

$\endgroup$

1 Answer 1

2
$\begingroup$

Let $f,g$ be both non-negative measurable functions and define: $$ E_n=\{x: g(x)\geq f(x)+t_n\}, $$ where $t_n$ is a sequence of decreasing positive numbers converging to $0$ as $n\to\infty$. For this choice, we have: $$ \mu(E_n)=\int_{E_n}d\mu\leq \int_{E_n}\frac{g(x)-f(x)}{t_n}d\mu=\int_{E_n}\frac{g(x)}{t_n}d\mu-\int_{E_n}\frac{f(x)}{t_n}d\mu\leq 0, $$ where we used the linearity of integration. To see why the linearity of integration holds note that $g(x)-f(x)$ is non-negative over $E_n$ and so are $g$ and $f$. You can define $h_n=(g-f)1_{g-f\geq t_n}$ and do the steps with $h_n$.

Therefore $\mu(E_n)$ is zero and so is $\mu(\cup E_n)=0$, implying that: $$ \mu(\{x: g(x)> f(x)\})=0. $$

$\endgroup$
5
  • $\begingroup$ I think I made a mistake in the question I asked. If we define $f,g:\Omega\to\mathbb{R}$ then $f$ and $g$ have to be integrable (i.e. be in $L^1(\mu)$) for $\int_A fd\mu$ to even make sense and then the linearity of the integral can be used for the second equality in your main line. What I MEANT to ask however, was what if $f,g:\Omega\to[0,+\infty]$ and so we allow the integrals to become $+\infty$ as well. Then what can we say? I don't think your proof still holds. I am sorry for the confusion. I will edit my question. $\endgroup$
    – RozaTh
    Sep 8, 2018 at 23:42
  • $\begingroup$ I see; but the answer is still true since $g-f$ is positive over $E_n$ and therefore the linearity still holds. I slightly changed the answer to be more clear. Note that the ambiguity around $\infty$-valued functions is implicitly addressed inside the order relation defined on functions and integral values, $\endgroup$
    – Arash
    Sep 9, 2018 at 19:10
  • $\begingroup$ @ Arash Sorry to pester you again, but I am still little unease about this. I totally agree that over $E_n$, $\frac{g(x)-f(x)}{t_n}$ is positive and its integral could become $\infty$. But both terms on the right side of the equality, that are $\int_{E_n}\frac{g(x)}{t_n}d\mu$ and $\int_{E_n}\frac{f(x)}{t_n}d\mu$, could become $+\infty$ as well, which make their subtractions an indeterminate form. In other words, from $\int_A fd\mu\geq \int_A gd\mu$ can we conclude $\int_A fd\mu-\int_A gd\mu\geq0$ when both terms can become $+\infty$? I appreciate further explanation. Thanks! $\endgroup$
    – RozaTh
    Sep 10, 2018 at 4:55
  • 1
    $\begingroup$ @RozaTh, This is why I mentioned the ambiguity around infinity valued functions in my comment. For infinity valued integrals you cannot make sense, in standard extended real functions, of the order between infinity values. How do you say $\int g\leq \int f$? So it is definitely true that $\int_{E_n}(g-f)d\mu+\int_{E_n}f d\mu$ is equal to $\int_{E_n}gd\mu$ and you are right to say the subtraction would not hold for infinity valued cases. But in that case, the inequality $\int g\leq \int f$ would be also problematic. $\endgroup$
    – Arash
    Sep 10, 2018 at 9:56
  • 1
    $\begingroup$ @RozaTh, one way to get around this issue would be to assume that $f$ and $g$ have finite integration over compact sets and then instead of $E_n$ using $E_n\cap C_n$ for compact sets $C_n\uparrow \Omega$. $\endgroup$
    – Arash
    Sep 10, 2018 at 10:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .