Theorem 7.34 in Apostol's MATHEMATICAL ANALYSIS, 2nd ed: Is this condition in the statement essential?

Question

Here is Theorem 7.34 (the Second Fundamental Theorem of Integral Calculus), in Chapter 7, in the book Mathematical Analysis - A Modern Approach To Advanced Calculus by Tom M. Apostol, 2nd edition:

Assume that $f$ is a real-valued, Riemann-integrable function on $[a, b]$. Let $g$ be a function defined on $[a, b]$ such that the derivative $g^\prime$ exists in $(a, b)$ and has the value $$ g^\prime (x) = f(x) \ \mbox{ for every $x$ in $(a, b)$}. $$ At the ednpoints assume that $g(a+)$ and $g(b-)$ exist and satisfy $$ g(a) - g(a+) = g(b) - g(b-).$$ Then we have $$ \int_a^b f(x) \, dx = \int_a^b g^\prime(x)\, dx = g(b) - g(a). $$

And, here is Apostol's proof of this theorem:

For every partition of $[a, b]$, we can write $$ g(b) - g(a) = \sum_{k=1}^n \left[ g \left( x_k \right) - g \left( x_{k-1} \right) \right] = \sum_{k=1}^n g^\prime \left( t_k \right) \Delta x_k = \sum_{k=1}^n f \left( t_k \right) \Delta x_k, $$ where $t_k$ is a point in $\left( x_{k-1}, x_k \right)$ determined by the Mean Value Theorem of differential calculus. But, for a given $\varepsilon > 0$, the partition can be taken so fine that $$ \left\lvert g(b) - g(a) - \int_a^b f(x) \, dx \right\rvert = \left\lvert \sum_{k=1}^n f \left( t_k \right) \Delta x_k - \int_a^b f(x) \, dx \right\rvert < \varepsilon, $$ and this proves the theorem.

Although I understand the statement as well as the proof of the Second Fundamental Theorem of Integral Calculus as given by Apostol, there is one catch.

What is the point of the very last equality in the statement of the theorem? For the Mean Value Theorem of Differential Calculus to be applicable, it should suffice for the function $g$ to be either continuous from the left at $a$ and from the right at $b$, or for the one sided limits $g(a+)$ and $g(b-)$ to exist as finite values.

Answer 1

The key is that we assume that $g$ is differentiable only on the open interval $(a, b) $ and not necessarily at end points. Further the equality $g'=f$ is also on the open interval $(a, b) $.

The assumption about existence of $g(a+),g(b-) $ is essentially equivalent to saying that $g$ is continuous on closed interval $[a,b] $. For, we can always redefine $g$ at end points $a,b$ to match its limits on those points and make it continuous.

The theorem then says that $$\int_a^b f(x) \, dx=g(b-) - g(a+)\tag{1} $$ The equality $$g(a) - g(a+) = g(b) - g(b-)$$ is only to replace the right hand side of equation $(1)$ with $g(b) - g(a) $.

The real surprise is that the assumption on existence of limits of $g$ at end points $a, b$ is unnecessary. Rather it can be proved that these limits exist using Riemann integrability of $f$.

More formally and more generally we can state the fundamental theorem as follows:

Second Fundamental Theorem of Calculus: Let $f:[a, b] \to\mathbb{R} $ be Riemann integrable on $[a, b] $ and let $g:(a, b) \to\mathbb{R} $ be a function such that $g'=f$ on $(a, b) $. Then the limits $g(a+), g(b-) $ exist and $\int_a^b f(x) \, dx=g(b-) - g(a+) $.

To prove it we first establish the result under the conditions that $g$ is defined and continuous on $[a, b] $ and $g'=f$ in $(a, b) $. This is exactly what is proved in the question using mean value theorem. And then we can prove the general version stated above by taking an arbitrary but fixed point $c$ in $(a, b) $ and note that for $c<x<b$ we have $$g(x) =\int_c^x f(t) \, dt+g(c) $$ Now letting $x\to b^-$ we get $$g(b-) =\int_c^b f(t) \, dt+g(c) \tag{2}$$ Similarly for $a<x<c$ we have $$g(c) - g(x) =\int_x^c f(t) \, dt$$ ie $$g(x) =g(c) - \int_x^c f(t) \, dt$$ Letting $x\to a^+$ we get $$g(a+) = g(c) - \int_a^c f(t) \, dt\tag{3}$$ Subtracting $(3)$ from $(2)$ we get $$g(b-) - g(a+) =\int_a^c f(t) \, dt+\int_c^b f(t) \, dt=\int_a^b f(t) \, dt$$ and we are done.