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$f(x),g(x):[0,1]\to [0,1]$,Prove that:$$\int_0^1f(g(x))dx\leqslant \int_0^1f(x)dx+\int_0^1g(x)dx$$ The solution uses $F(x)=f(x)-x$ and turns the problem into:$$\int_0^1F(g(x))dx\leqslant \frac12$$ Is there more obvious or essential approach?

Jam
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Alpha Beta
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    https://math.stackexchange.com/q/276722/42969, https://math.stackexchange.com/q/4033497/42969 – Martin R Sep 25 '22 at 16:00
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    In Martin R.'s link, the statement is proved for $f$ monotone increasing, and indeed, the statement is not necessarily true if $f$ does not satisfy this condition. Continuity conditions on $f,g$ may also be necessary. – legionwhale Sep 25 '22 at 16:10
  • $g \equiv 0$, $f(x)=1_{{0}}(x)$. – PhoemueX Sep 25 '22 at 18:26

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