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I'm trying to prove:

$f,g$ integrable, then $\max \{f,g\}$ and $\min\{f,g\}$ are integrable. Since $f$ is integrable, then, by this definition of integrability:

$$S(f,P)-s(f,P) <\epsilon$$ *same for $g$.

I'll name $\phi = \max\{f,g\}$ and $\psi = \min\{f, g\}$

We have to prove that:

$$S(\phi, P)-s(\phi, P)<\epsilon$$

where:

$$S(\phi, P) = \sum M_i(t_{i}-t_{i-1})$$

Being $M_i = \sup{\phi(x)} = \sup{\max\{f(x), g(x)\}}$ in the interval $[t_{i-1}, t_i]$

and the similar definition for $s(\phi, P)$

The question is: what does the $\sup$ of the $\max$ even means?

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Guerlando OCs
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    Hint: $$max{f(x),g(x)}=\frac{|f(x)-g(x)|+(f(x)+g(x))}{2}$$ and $$min{f(x),g(x)}=\frac{f(x)+g(x)-|f(x)-g(x)|}{2}.$$ – Rafael Aug 25 '16 at 00:55

1 Answers1

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You can express $\max\{ f,g \}$ as $$ \frac{f+g + |f-g|}{2} $$ Then $\sup$ of the $\max$ would be the $\sup$ of this function. However, better than resorting to the $\epsilon$ definition is to prove that $\frac{f+g+|f-g|}{2}$ is integrable whenever $f$ and $g$ are. You could do this by proving that the sum (and difference) of integrable functions is integrable and that the absolute value of an integrable function is integrable.

Shai
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