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Here $u=u(x,y)$ is a two-variable integrable function defined on $D\subset\mathbb{R}^2$, $f(x)$ is a one-variable function defined on $\mathbb{R}$.

What is trival is that if $f(x)$ is continuous, then $f(u(x,y))$ is integrable on $D$.

Now $f(x)$ is just integrable, is $f(u(x,y))$ still integrable on $D$?

YZ Zhang
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  • Modify the answer given here https://math.stackexchange.com/questions/1060834/composition-of-two-riemann-integrable-functions – Reveillark Jan 12 '20 at 04:21

2 Answers2

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It's not true that $f(u(x,y))$ is integrable if $f$ is continuous (even if it is also integrable).

Take $ f(x) = 1/(1+x^2)$, which is integrable and continuous on $\mathbb R$.

$u(x,y) = 0$ is integrable on $\mathbb R^2$. $f(u(x,y)) = 1$ is not integrable there.

Robert Israel
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I don't have a tangible answer but I am imagining something like this. $f$ is a function with discontinuities over set of measure zero $\Omega$, i.e. $f(x^+)\neq f(x^-)$ if $x\in\Omega$. $u(x,y)$ is a function such that if $x,y\in S$ with $S$ subset (of non-zero measure) of $D$, $u(x,y)\in D$. This would be a counter-example.

Abdullah Ali Sivas
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