Computing Riemann integral (if you know the integral of a similar function)

Question

Suppose I have a function f, that differs from another function g by exactly one point, let's say at c.

I know that I can conclude that f is Riemann integrable on [a,b] if g is on the same interval. Now, my question is : so we have c in [a,b] and we know that the Riemann integral of g on our interval is 0, can I conclude the same thing for f ?

I know other ways to work with Riemann integrability such as the squeeze theorem, but I want to know if this would be a valid shortcut in our situation.

Thanks !

Answer 1

Yes. If $f$ and $g$ differ only on a set of measure $0$, which any finite set is, then $\int fdx=\int gdx$. Since $f$ and $g$ differ only at a single point $c$ in your example, the upper and lower sums of $f$ and $g$ will be the same and so their integral will be the same if $g$ is Riemann integrable.

More detailed version: Let $g$ be a Riemann integrable function on $[a,b]$ and let $f$ be a function defined on $[a,b]$ with $g(x)=f(x)\quad \forall x\in[a,c)\cup(c,b]$. Since $g$ is Riemann integrable, $U(g)=L(g)=\int_a^b g(x)dx$ is defined. Since $f$ and $g$ differ only at a single point, $U(g)=U(f)=L(g)=L(f)$. Thus, $\int_a^b f(x)dx=U(f)=U(g)=\int_a^bg(x)dx$. So $f$ is Riemann integrable and $\int_a^bf(x)dx=\int_a^bg(x)dx$.

Answer 2

If $f$ differs from $g$ at only one point then you have that $f(x)-g(x)=0$ at all except one point so you have $\int_{a}^{b}(f(x)-g(x))dx=0$ so their integrals are the same number because of $\int_{a}^{b}(f(x)-g(x))dx=0=\int_{a}^{b} f(x)dx - \int_{a}^{b} g(x)dx$ so we have $\int_{a}^{b} f(x)dx=\int_{a}^{b} g(x)dx$.

You could use exactly the same argument if $f$ differs from $g$ at some finite number of points, but if they differ in an infinite number of points the situation may become different, depending on whether the infinite number of points is countable or uncountable set.