# Proof that improper integration works

I wasn't really sure how to title this, but the way the problem is phrased makes me think of improper integrals so I thought that might be a good title.

Let $f\colon[a,b]\to\mathbb{R}, a <b,$ be Riemann integrable on $[a,b]$. Show that $$\int_a^b f(x) dx = \lim_{T\to b}\int_a^T f(x)dx$$ I don't really see that there's a whole lot to do here, which is part of my confusion; it feels like it should be harder than this. Let $F$ denote the antiderivative of $f$ which exists by assumption, then apply the fundamental theorem of calculus: $$\lim_{T\to b}\int_a^T f(x)dx = \lim_{T\to b } F(T)-F(a) = F(b)-F(a) = \int_a^b f(x) dx$$ The only thing that feels a little shaky is justifying $\displaystyle\lim_{T\to b } F(T)=F(b)$. We're only given that $f$ is Riemann integrable, which says nothing about the continuity of the function. What prevents something weird happening at $b$?

Edit: The fundamental theorem on calculus, Rudin version: If $f$ is Riemann integrable on $[a,b]$ and if there is a differentiable function $F$ on $[a,b]$ such that $F'=f$, then $$\int_a^b f(x) dx = F(b)-F(a)$$

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Part of the fundamental theorem of calculus is that $F$ is continuous if $f$ is Riemann integrable. What is your version of the fundamental theorem of calculus? – lhf Oct 23 '12 at 1:14
One that I apparently haven't read completely in a while. I'll go read it again. – chris Oct 23 '12 at 1:18
Just to make sure that I completely understand what you're saying, by knowing that $f$ is Riemann integrable, we're guaranteed a continuous antiderivative? The version in my calculus textbook only makes an assumption that $f$ is continuous which I don't know to be the case here. What if $f$ is piecewise continuous instead? How do I know what $F$ doesn't have a discontinuity at $b$ (assuming there is one at $f$)? – chris Oct 23 '12 at 1:25
Okay. I think I realized where I went wrong. We have specifically that $F'=f$, which implies that we have a differentiable function. But differentiability implies continuity, so $F$ must be continuous, and so we get that $\lim_{T\to b} F(T)=F(b)$. – chris Oct 23 '12 at 1:28
No, that's not it. Define $F(x)=\int_a^x f(t) \, dt$. If $f$ is integrable, then the FTC tells you that $F$ is continuous. If $f$ is continuous, then the FTC tells you that $F$ is differentiable. You are not granted this second assumption here. As @lhf pointed out, this depends on the version of the FTC that you learned. – wj32 Oct 23 '12 at 1:56

6.20 Theorem Let $f$ be Riemann integrable on $[a,b]$. For $a \le x \le b$, put $$F(x)=\int_a^x f(t) \, dt.$$ Then $F$ is continuous on $[a,b]$; furthermore, if $f$ is continuous at a point $x_0$ of $[a,b]$, then $F$ is differentiable at $x_0$, and $$F'(x_0)=f(x_0).$$
You have already stated the other theorem (6.21) in your edit, but you do not need that theorem to answer your question. (In fact, it does not apply here since $f$ may not be continuous.) If $f$ is Riemann integrable then $F$ (as defined in Theorem 6.20) is continuous on $[a,b]$. Then $\lim_{x \rightarrow b} F(x) = F(b)$, i.e. $$\lim_{x \rightarrow b} \int_a^x f(t) \, dt = \int_a^b f(t) \, dt.$$
Your proof states "Let $F$ denote the antiderivative of $f$ which exists by assumption", which is simply not true!