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If f(x) and g(x) are Riemann integrable in [a,b], why $h(x)=\max\{f(x),g(x)\}$ is still Riemann integrable in [a,b]?

Or maybe it is wrong?

Martin Sleziak
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    Note that $\max(f,g)=(f+g+\lvert f-g\rvert)/2$. – Yai0Phah Apr 19 '14 at 04:13
  • See http://math.stackexchange.com/questions/86031/if-f-and-g-integrable-then-max-f-g-is and http://math.stackexchange.com/questions/72844/is-the-pointwise-maximum-of-two-riemann-integrable-functions-riemann-integrable (BTW the list of related question on the right is often useful. One of these two questions is shown there.) – Martin Sleziak Apr 19 '14 at 07:19

1 Answers1

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Yes, the $Max$ function of two Riemann -integrable functions is Riemann -integrable. This is because the $Max$ of two a.e -continuous bounded functions is also a.e -continuous and bounded.

For $f,g$ continuous, the function $Max${f,g} is continuous. This implies that (since a Riemann-integral must be a.e -continuous) , that $Max${f,g} is a.e -continuous; the points of discontinuity are contained in the union of the points of discontinuity of $f,g$.

Maybe a good (informal) way of seeing that the Max function is continuous for $f,g$ continuous (and then adjust for the fact that they are a.e. continuous) is to draw the graph of two continuous functions $f,g$ and look at the only possible problem points, i.e., the points {$x_i$} where the functions go from $f(x_i)<g(x_i)$ to $f(x_i)=g(x_i)$ to $f(x_i)>g(x_i)$; take a neighborhood left of $x_i$ and one right of it. Then do a $\delta - \epsilon$ argument for $g$ at the left of $x_i$, and one for $f$ at the right of $x_i$ to show the $Max$ function is continuous at those points. At every other point, $Max$ will equal either $f$ or $g$ , both of which are continuous. A similar argument works for the boundedness of $Max${f,g}.

Martin Sleziak
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