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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.

t.b.
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August
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2 Answers2

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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 $[a,b]$.

alejopelaez
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  • or simply that $\left|\max(f(x),g(x))\right|\le|f(x)|+|g(x)|$ – robjohn Oct 15 '11 at 18:48
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    @robjohn, the fact that $|h(x)|\leqslant k(x)$ with $k$ Riemann integrable does not imply that $h$ is Riemann integrable. At all. – Did Oct 15 '11 at 22:10
  • @alejo: Sorry but I am not sure to see how your hint applies. – Did Oct 15 '11 at 22:11
  • @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. – robjohn Oct 15 '11 at 22:53
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    @Didier: Well, if $f$ if Riemann integrable, so its $|f|$ and any linear combination of Riemann integrable functions is also Riemann integrable. – alejopelaez Oct 15 '11 at 23:35
  • @alejopelaez How can i prove that $$ \left| f \right| $$ is also Riemann integrable? This result is not true in any kind of domains, for example in a infinite interval it is not true. But if the interval is finite? or in the best of the cases, is also closed, how can i prove it? – August Oct 16 '11 at 19:45
  • @August: I added some more info in my answer that should help you prove that $|f|$ is also Riemann integrable. Also keep in mind that Riemann integrability is defined for bounded functions on closed intervals. For other type of intervals you define improper Riemann integral, for which I'm not sure if this holds. – alejopelaez Oct 16 '11 at 21:29
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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.

Did
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