Proof that $F$ is continuous in fundamental theorem of calculus

Question

I am studying calculus and have arrived at the first FTC. In our book it states that:

let $f: [a,b]\to\mathbb{R}$ be continuous on the interval $[a,b]$ then we define the function $F_f : [a,b]\to\mathbb{R} : x \mapsto F_f(x) := \int_a^x f(t)\,dt$

Then these properties hold:

• $F_f$ is continuous on $[a,b]$
• $F_f$ is differentiable on $]a,b[$
• $F_f$ is a primitive for $f$

The proof of the last two properties are quite understandable and also easy to find. However I am confused about the first property (supposedly the most straightforward one) where it states:

The continuity of $F_f$ follows immediately from the fact that for every $x,y \in [a,b]$ we have:

$|F_f(x) - F_f(y)| = \int_a^x f(t)\,dt - \int_a^y f(t)\,dt| = |\int_y^x f(t)\,dt| \leq \max_{x \in [a,b]} |f(x)|\cdot |x-y|$

Now here is what I am not sure of:

• Where it says $\max_{x \in [a,b]} |f(x)|$ are we talking about a different $x$ here? Like it could have been $\max_{t \in [a,b]} |f(t)|$, right?

• This statement seems true enough and understandable (assuming the above) however: how does continuity of $F_f$ follow from it?

Answer 1

• Right.
• Call it $M$. That is, let $M=\sup_{x\in[a,b]}\bigl\lvert f(x)\bigr\rvert$. Then$$(\forall x,y\in[a,b]):\bigl\lvert F_f(x)-F_f(y)\bigr\rvert\leqslant M\lvert x-y\rvert.$$So, for any $\varepsilon>0$, if you take $\delta=\frac\varepsilon M$, then$$\lvert x-y\rvert<\delta=\frac\varepsilon M\implies\bigl\lvert F_f(x)-F_f(y)\bigr\rvert\leqslant M\lvert x-y\rvert<M\frac\varepsilon M=\varepsilon.$$

Answer 2

$F$ is continuous if

$\lim_\limits{x\to y} F(x) = F(y)$ for all $y$ in $[a,b]$

$\forall\epsilon>0,\exists \delta >0,\forall x,y \in [a,b]: |x-y|<\delta \implies |F(x) - F(y)|<\epsilon$

$F(x) - F(y) = \int_a^x f(x) \ dx - \int_a^y f(x) \ dx = \int_y^x f(x) \ dx$

$f(x)$ is bounded over the interval, and has a least upper bound. Let $M$ be the least upper bound of $|f(x)|$

$M = \sup \{|f(x)|: x\in[x,y]\}$

$|\int_y^x f(x) \ dx| \le |x-y|M$

$\delta = \frac {\epsilon}{M}$

$F(x)$ is continuous.