Proving Continuity of F(x) Without the Fundamental Theorem of Calculus

[Vespero]

Homework Statement

Without using the Fundamental Theorem of Calculus:

Let f be continuous on the compact interval [a,b].
Show that F(x) = ∫f(t)dt from a to x.

Homework Equations

We know that if f is continuous on [a,b], then f is integrable.
If a function is differentiable, it is continuous.

The Attempt at a Solution

I think I am just being blinded by not being able to use the FTC, and that this is a fairly simple problem. I am just not sure exactly where to start. If I could show that F(x) was differentiable, then it must be continuous, but I can't think of a way to do that without the FTC. Any help would be greatly appreciated.

 

[Office_Shredder]

The FToC is overkill here. F(x) is continuous if and only if F(a+h)-F(a) goes to zero as h goes to zero for each a. Write down what F(a+h) and F(a) are, and try to find an upper bound for F(a+h)-F(a)

 

[fzero]

I'd try to use

F(x) - F(c) = \int_c^x f(t) dt

in the \epsilon-\delta definition. You don't exactly need the FTC, but you could use either the mean-value theorem or maybe just a crude Riemannian approximation to the value of the integral.

 

[Vespero]

That's what I ended up doing. I forgot we covered several of the properties of integrals that make that statement valid. It allowed me to show that |F(c)-F(x)| <= M(c-x), which causes |F(c)-F(x)| to converge to 0 as x approaches c.

 

[Vespero]

The entirety of the proof as I have written it is as follows:

Since f(x) is continuous on [a,b], then f(x) is integrable on [a,b] (Theorem.) Since f(x) is integrable on [a,b], f(x) is bounded (Theorem.) That is, there exists a number M such that |f(x)| <= M.

Choosing a value x in [a,b],
|F(c)-F(x)| = |int(a,c)f(t)dt - int(a,x)f(t)dt| <= int(x,c)|f(t)|dt <= int(x,c)Mdt = M(c-x)

Assuming c > x and taking the limit as x approaches c, we see that
lim(x-->c) |F(c)-F(x)| <= lim(x-->c) M(c-x) = 0

But |F(c)-F(x)| must be non-negative, so
lim(x-->c) |F(c)-F(x)| = 0.

Therefore, F(x) = int(a,x)f(t)dt is continuous at every point x in [a,b] and is thus continuous on [a,b] by the definition of continuity on an interval.
If you see anything that doesn't follow or that is superfluous, could you point it out or give a suggestion as to a better way to have the proof flow?

 

[Office_Shredder]

Since f(x) is integrable on [a,b], f(x) is bounded (Theorem.)

This isn't true. For example 1/sqrt(x) is integrable on [0,1]. f(x) is bounded on [a,b] because f(x) is continuous

Choosing a value x in [a,b],
|F(c)-F(x)| = |int(a,c)f(t)dt - int(a,x)f(t)dt| <= int(x,c)|f(t)|dt <= int(x,c)Mdt = M(c-x)

It really should be M|c-x| here since you don't know which is larger. You have to be a little bit careful when you write int(x,c)|f(t)|dt because you might have just said a non-negative number is smaller than a negative number. It's best to leave the absolute value signs on the outside as well
|F(c)-F(x)|\leq |\int_x^c |f(t)|dt | \leq |\int_x^c M dt| = |M(c-x)|

 

[Vespero]

This isn't true. For example 1/sqrt(x) is integrable on [0,1]. f(x) is bounded on [a,b] because f(x) is continuous.

By integrable, I mean Riemann integrable, which is what we are using at the moment. 1/sqrt(x) isn't Riemann integrable on [0,1], as it isn't continuous and bounded on that interval. However, f(x) is continuous on [a,b], so it is Riemann integrable, which implies that it is bounded. I should have said Riemann integrable. Is this correct?

It really should be M|c-x| here since you don't know which is larger. You have to be a little bit careful when you write int(x,c)|f(t)|dt because you might have just said a non-negative number is smaller than a negative number. It's best to leave the absolute value signs on the outside as well
|F(c)-F(x)|\leq |\int_x^c |f(t)|dt | \leq |\int_x^c M dt| = |M(c-x)|

I do agree here. I should have been paying more attention to that. Thanks.