If f is integrable and g is bounded in [a,b] then the upper integral of f+g = integral of f+ upper integral of g

Asked Apr 07 '21 at 17:55

Active Apr 10 '21 at 10:12

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I need to prove the following when f is integrable:

$\overline{\int_{a}^{b}}(f+g)(x)dx = {\int_{a}^{b}}f(x)dx+\overline{\int_{a}^{b}}g(x)dx$

I'm not sure how to go about this proof, in the first part of the question I proved that if both are bounded then:

$\overline{\int_{a}^{b}}(f+g)(x)dx \le \overline{\int_{a}^{b}}f(x)dx+\overline{\int_{a}^{b}}g(x)dx$

but I'm not sure how to use f being integrable to prove these are equal. I do know that f being integrable also means I proved the $\le$ part of the question but how do I prove the other way?

Will appreciate any hint/tips...

edited Apr 10 '21 at 10:12

Ram Elgov

asked Apr 07 '21 at 17:55

lia

The integral of f is without an upper bar (when it's integrable) – lia Apr 07 '21 at 18:01
1

Denote by $U(f)$ and $L(f)$ the upper and lower integrals of $f$. You've proved that $U(f+g) \le U(f) + U(g)$, and a similar proof shows that $L(f) + L(g) \le L(f+g)$. Can you prove similarly that $$L(f) + U(g) \le U(f+g), ?$$ – Umberto P. Apr 07 '21 at 18:09
Thank you for your help, just to be sure I proved it correctly, it's right to say that: $sup L(f) + sup L(g)<= inf U(f+g)$ because f is integragble after I proved the above? – lia Apr 07 '21 at 18:33
I'm not sure what you mean by $\sup L(f)$. – Umberto P. Apr 07 '21 at 18:42
Ok I got confused but I can't understand how to prove this: $L(f) + U(g) <= U(f+g)$ it's not similiar to the way you prove the other two... – lia Apr 07 '21 at 19:36
You can look at this question which is about upper integrals: Additivity of definite integrals. And also at this recent question: Prove: $\overline{\int_{a}^{b}} (g+f)(x) \ dx = \int\limits_a^b {{f(x)}dx}+ \overline{\int_{a}^{b}} g(x) \ dx$. – Martin Sleziak Apr 21 '21 at 06:55

If f is integrable and g is bounded in [a,b] then the upper integral of f+g = integral of f+ upper integral of g

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