elementary prove or disprove question about integrability Let $f:[a,b] \longrightarrow R$ integratable in that interval. Let $g:[a,b] \longrightarrow R$
s.t. there is a point $c \in [a,b]$ that $\forall x \ne c$ $g(x) = f(x)$
and if $x = c$ then $g(c) = 1 + f(c)$. Then $g$ is integratable in that interval
and $\int_{a}^{b} f(t) dt = \int_{a}^{b} g(t) dt$.
My thoughts:
I think the statement is correct. Intuitively, I know that the antiderivative is defined up to a constant( I mean that $ F + c = F$) because when you differentiate it you remain with $F'$. And that $g(x) = f(x)$ when $f$ is  integratable except for one point($c$). which Intuitively means that g has a finite number of non-continuous points which means that $g$ is still integratable. As you can see, all that I'm saying is very intuitive and not formal at all. 
Can someone help me to formalize it please?
Thanks in advance!
 A: ntt is right. A bounded function $f:[a,b]\to\mathbb{R}$ is Riemann integrable if and only if its set of discontinuity points has Lebesgue measure zero. So changing the definition of $f$ at one point does not destroy integrability, since a set of Lebesgue measure zero union a singleton still has Lebesgue measure zero. Also the integral does not change. To see this, you have to use the definition of upper and lower Riemann integral. Given any partition $P$ given $t_0=a<\ldots<t_n=b$ you can always add more points to the partition, so you can always assume that $c$, $c-\delta$ and $c+\delta$ are points of the partition. Now,
$$\vert\delta\inf_{[c-\delta,c]}f-\delta\inf_{[c-\delta,c]}g\vert\le \delta ( M+1),\quad \vert\delta\sup_{[c-\delta,c]}f-\delta\sup_{[c-\delta,c]}g\vert\le \delta ( M+1),$$ where $M$ is the bound for $|f|$. Same in $[c,c+\delta$, and so the lower sum of $f$ and the lower sum of $g$ over $P$ differ at most by $2\delta ( M+1)$. Same for upper sum. So the lower and upper integral of $f$ and $g$ coincide. I am skipping some details.
