Use of Continuity in the Fundamental Theorem of Calculus proof I wanted to check my understanding of the authors use of continuity in this proof of the FTC: $A'(t) = \lim_{h \to 0}\frac{A(t+h)-A(t)}{h} = f(t)$
Context
Using our classic approach of defining a derivative through limits, where $A(x)$ is an area function i.e. integral, my text book gets:
$\lim_{h \to 0^{+}} f(t + h) \le  \lim_{h \to 0^{+}}\frac{A(t+h)-A(t)}{h} \le \lim_{h \to 0^{+}}f(t)$
$f(t) \le  \lim_{h \to 0^{+}}\frac{A(t+h)-A(t)}{h} \le f(t)$
$\lim_{h \to 0^{+}}\frac{A(t+h)-A(t)}{h} = f(t)$
Where we have assumed $h > 0$ and we can use a similar argument for $h < 0$, combining the two gives us the required theorem:
$A'(t) = \lim_{h \to 0}\frac{A(t+h)-A(t)}{h} = f(t)$
Question
The author notes that continuity is essential to justify line 1 to line 2, i.e. that:
$\lim_{h \to 0^{+}} f(t + h) = f(t)$
Is the reason as follows: Only if a function is continuous do we have:
$\lim_{x \to c} f(x) = f(c)$
So letting $x = t + h$
$\lim_{x \to t} f(x) = \lim_{h \to 0^{+}} f(t + h) = f(t)$ only holds if $f$ is continuous? Where we have used $h \to 0^{+}$ because we assume $h > 0$?
Intuitively could i say something like this: We can't assume that $ x \in [t,t + h]$ goes to t from $t + h$, as $h$ get's smaller and smaller, unless the function is continuous over this interval, and as h could take any possitive value the function must be continuous $\forall x \ge t$...or would it be $ \forall h \ge 0$?
I don't think this 'intuitive' explenation is great, and I realise i'm unclear whether the function needs to be continuous beyond $t$ or beyond $0$ for this to work? Thoughts greatly appreicated thanks!
 A: For definiteness, let's agree that $t_{0}$ and $t$ are real numbers, $f$ is a Riemann-integrable function on some open interval $I$ containing $t_{0}$ and $t$, and
$$
A(t) = \int_{t_{0}}^{t} f(s)\, ds.
$$
Consider the three conditions:

*

*$f$ is continuous at $t$.

*$\lim\limits_{h\to 0^{+}} f(t + h) = f(t)$.

*$A$ is differentiable at $t$, and $A'(t) = f(t)$.

Loosely, a function $f$ is continuous at $t$ if and only if both one-sided limits at $t$ exist and are equal to $f(t)$. That is, condition 1 implies condition 2; conversely, condition 2 and the analogous condition for a left-hand limit imply condition 1.
Further, condition 1 implies condition 3 (this is part of the FTC), but condition 3 does not imply condition 1. A typical counterexample is to let $f(s) = 1$ if $s = 1/n$ for some non-zero integer $n$, and $f(s) = 0$ otherwise. The integral $A$ is identically $0$ (!), hence differentiable, and $A'(0) = 0 = f(0)$ (condition 3 holds) but the function $f$ is discontinuous at $0$ (condition 1 fails). In this sense, it would be inaccurate to say "continuity of $f$ at $t$ is essential to deduce that $A'(t) = f(t)$." (That's not your authors' claim; just saying.)

Next let's examine "line 1" in the argument: If $t + h \in I$, then
$$
\frac{A(t+h) - A(t)}{h}
= \frac{1}{h}\biggr[\int_{t_{0}}^{t+h} f(s)\, ds - \int_{t_{0}}^{t} f(s)\, ds\biggr]
= \frac{1}{h} \int_{t}^{t+h} f(s)\, ds.
$$
In general, this expression is neither bounded below by $f(t + h)$ nor bounded above by $f(t)$. In this sense, I don't understand where
$$
\lim_{h \to 0^{+}} f(t + h)
\leq \lim_{h \to 0^{+}}\frac{A(t + h) - A(t)}{h}
\leq \lim_{h \to 0^{+}}f(t)
$$
comes from. (If $f$ is continuous at $t$, or even continuous from the right at $t$, the preceding inequality is true, but to deduce it we'd effectively need the theorem we're attempting to prove.)
Be that as it may, if the preceding inequality is established, then the limits do evaluate as in line 2 because condition 1 implies condition 2. But based on the context given, it's difficult to evaluate the claim continuity is essential to justify line 1 to line 2: If $f$ is not continuous at $t$, the leftmost limit in line 1 may or may not exist, separately from whether the leftmost limit is equal to $f(t)$. But this does not appear to be the pedagogical point being made.

Finally, though this was not asked, here's how I'd prove the FTC, avoiding the points in question here: If $t + h \in I$, then
\begin{align*}
  \biggl|\frac{A(t + h) - A(t)}{h} - f(t)\biggr|
  &= \biggl|\frac{1}{h} \int_{t}^{t+h} f(s)\, ds - f(t)\biggr|
  && \text{preceding calculation} \\
  &= \biggl|\frac{1}{h} \int_{t}^{t+h} f(s)\, ds - \frac{1}{h} \int_{t}^{t+h} f(t)\, ds\biggr|
  && \text{$f(t)$ is constant} \\
  &= \biggl|\frac{1}{h} \int_{t}^{t+h} [f(s) - f(t)]\, ds\biggr|
  && \text{linearity of the integral} \\
  &\leq \biggl|\frac{1}{h} \int_{t}^{t+h} |f(s) - f(t)|\, ds\biggr|
  && \text{triangle inequality for integrals} \\
  &\leq \sup_{|s - t| \leq |h|} |f(s) - f(t)|
  && \text{monotonicity of the integral.}
\end{align*}
Because $f$ is continuous at $t$, the right-hand side can be made as small as we like by taking $|h|$ sufficiently small. By definition,
$$
A'(t) = \lim_{h \to 0} \frac{A(t + h) - A(t)}{h} = f(t).
$$
A: The following theorem, taken from the seventh chapter of Rudin's Real and Complex Analysis (third edition) shows that, in order for the indefinite integral to be differentiable at every point and its derivative equal the integrated function, it is  not essential that the integrated function is continuous. So you can take a differentiable function whose derivative is integrable but has discontinuities (by Darboux's theorem, necessarily discontinuities of the second kind, i.e. at least one of the one-sided limits doesn't exist).
]1
A: Let's see what happens with a discontinuous function. Consider, for example,
$$f:\mathbb{R}\to\mathbb{R},\quad x\mapsto\begin{cases}
1,&x=a,\\
0,&x\neq a,
\end{cases}$$
where $a\in\mathbb{R}$. Then
$$\lim_{h\to 0}f(a+h)=\lim_{h\to0}0=0,$$
but
$$f(a)=1,$$
and so
$$\lim_{h\to0}f(a+h)\neq f(a).$$
If we have a continuous function then your reasoning seems to work. Do you see now why continuity is required?
