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As the question states, if function $f$ and $g$ aren't Riemann integrable, can the sum of the functions be Riemann integrable?

Most importantly, in what way should this be proven?

Dole
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2 Answers2

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The answer is yes, take $h$ a non Riemann integrable function over some $[a,b]$, we have that $-h$ is also non Riemann integrable over $[a,b]$ but $$ -h+h=0 $$which is Riemann integrable over $[a,b]$.

Olivier Oloa
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The most obvious example has already been given: consider any non-Riemann integrable function and its oppossite. Here's a concrete example based on the same idea, consider the indicator functions of the rationals and the irrationals (e.g. on $[0,1]$): $$f(x) = \begin{cases} 1 & x\in\Bbb Q \\[2ex] 0 & x\notin\Bbb Q \end{cases} \quad\quad \mbox{and} \quad\quad g(x) = \begin{cases} 0 & x\in\Bbb Q \\[2ex] 1 & x\notin\Bbb Q \end{cases}$$ Then $f(x)+g(x) = 1$, clearly Riemann integrable.

See also: Is the indicator function of the rationals Riemann integrable?

Community
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StackTD
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    Though that's just the negation idea but with an added constant, you have g = -f + 1 – ilkkachu Apr 11 '17 at 07:22
  • @ilkkachu You're right of course, but I thought a concrete example could still be useful. I rewrote it a bit to reflect your (good) point. – StackTD Apr 11 '17 at 07:26