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Let $f,g$ be Riemann integrable functions, prove that the function $ h(x) $ defined by $$ h\left( x \right) = \max \left\{ {f\left( x \right),g\left( x \right)} \right\} $$ is also Riemann integrable.

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2 Answers 2

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Hint
Use the fact that $\max(f(x), g(x)) = \frac{f(x)+g(x)+|f(x)-g(x)|}{2}$
Edit
To use this fact we need to prove that if $f$ is Riemann integrable, so is $|f|$. For this we set for some interval $A$ where $f$ is bounded: $$ M = \sup{\{f(x) : x \in A\}}$$ $$ m = \inf{ \{f(x) : x \in A\} }$$ $$ M' = \sup{\{|f(x)| : x \in A\}}$$ $$ m' = \inf{ \{|f(x)| : x \in A\} }$$ And I will let you prove that $M' - m' \leq M - m$.
With this in mind, if $f$ is Riemann integrable on $[a,b]$ then take some partition $P$ of $[a,b]$, and using the previous fact prove that $$0 \leq U(|f|, P) - L(|f|, P) \leq U(f, P) - L(f, P)$$ Where U and L are the upper and lower sums respectively.
This will allow you to use the Riemann integrability criterion and conclude that $|f|$ is also integrable on $[a,b]$.

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  • $\begingroup$ or simply that $\left|\max(f(x),g(x))\right|\le|f(x)|+|g(x)|$ $\endgroup$
    – robjohn
    Commented Oct 15, 2011 at 18:48
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    $\begingroup$ @robjohn, the fact that $|h(x)|\leqslant k(x)$ with $k$ Riemann integrable does not imply that $h$ is Riemann integrable. At all. $\endgroup$
    – Did
    Commented Oct 15, 2011 at 22:10
  • $\begingroup$ @alejo: Sorry but I am not sure to see how your hint applies. $\endgroup$
    – Did
    Commented Oct 15, 2011 at 22:11
  • $\begingroup$ @Didier: Indeed. I was simply looking at alejo's answer and started thinking Lebesgue integral and this simplifies his answer in that case. We need to use boundedness and discontinuities of measure $0$ instead. $\endgroup$
    – robjohn
    Commented Oct 15, 2011 at 22:53
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    $\begingroup$ @Didier: Well, if $f$ if Riemann integrable, so its $|f|$ and any linear combination of Riemann integrable functions is also Riemann integrable. $\endgroup$ Commented Oct 15, 2011 at 23:35
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Call $\Delta_\sigma(u)$ the difference between the upper and lower Darboux sums of a function $u$ with respect to a subdivision $\sigma$. By definition of Riemann integrability, $\inf\limits_\sigma\Delta_\sigma(f)=\inf\limits_\sigma\Delta_\sigma(g)=0$. Call $M_I(u)$ and $m_I(u)$ the supremum and the infinum of a function $u$ on an interval $I$.

Then, for every interval $I$, $M_I(h)=\max\{M_I(f),M_I(g)\}$ and $m_I(h)\geqslant\max\{m_I(f),m_I(g)\}$, hence $$ M_I(h)-m_I(h)\leqslant\max\{M_I(f)-m_I(f),M_I(g)-m_I(g)\}, $$ which implies $$ M_I(h)-m_I(h)\leqslant M_I(f)-m_I(f)+M_I(g)-m_I(g). $$ Summing this over the intervals $I$ defining a subdivision $\sigma$, one gets $\Delta_\sigma(h)\leqslant\Delta_\sigma(f)+\Delta_\sigma(g)$.

For every positive $t$, there exists $\sigma$ such that $\Delta_\sigma(f)\leqslant t$ and $\tau$ such that $\Delta_\tau(g)\leqslant t$. For every subdivision $\varrho$ containing $\sigma$ and $\tau$, one gets $$ \Delta_\varrho(h)\leqslant \Delta_\varrho(f)+\Delta_\varrho(g)\leqslant\Delta_\sigma(f)+\Delta_\tau(g)\leqslant2t. $$ This proves that $\inf\limits_\sigma\Delta_\sigma(h)=0$ hence $h$ is Riemann integrable.

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