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The problem states: Let $f(x),g(x)$ be Riemann integrable over $[a,b]$. Define $h:[a,b]\rightarrow\mathbb{R}$ as: $$h(x)=\max(f(x),g(x))$$

Prove that $h(x)$ is Riemann integrable over [a,b]

-I know that because $f$ and $g$ are integrable, then:

$\forall \epsilon>0$, there exists a partition P so that: $U(f,P)-L(f,P)<\epsilon$

$\forall \epsilon\prime>0$, there exists a partition P$\prime$ so that: $U(g,P\prime)-L(g,P\prime)<\epsilon\prime$

I also know that: $maxf(x)\le\sup f(x)$ and $maxg(x)\le\sup g(x)$ where both functions are defined.

Any hints on how should Iaproach this proof? Thanks

BigM
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mobzopi
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  • This can be done easily if you can make use of "a function is Riemann integrable over $[a,b]$ iff its set of discontinuities has measure zero", for if $f$ and $g$ are Riemann integrable, then its respective discontinuities sets have measure zero, and the set of discontinuities of $h$ will be a subset of the union of the mentioned above plus a measure zero set. – Daniel Feb 10 '15 at 03:33
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    I'm not allowed to use that theorem for now(since we have not proven it in class) – mobzopi Feb 10 '15 at 03:36
  • Possible duplicate of http://math.stackexchange.com/q/72844/148510 – RRL Feb 10 '15 at 04:08

1 Answers1

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You can use the fact that for any two $a,b \in \mathbb R$, we have the identity $$(*)\quad \max(a,b) = \frac{a+b+|a-b|}{2} $$ to say that $$ \max(f(x),g(x)) = \frac{f(x) + g(x) + |f(x)-g(x)|}{2}. $$ The absolute value, the sum, and the difference, and the multiplication by a constant of Riemann integrable functions are all Riemann integrable, hence the max is also integrable.

Proof of $(*)$ :

If $a\ge b$, we have that |a-b| = a-b and the identity is $\frac{a+b + a - b}{2} = a $, which is the max. If $a< b$, then $|a-b| = b-a$ and the computation is similar.

Jean Lagacé
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