Proving the Limit of Integral over f:[a,b] to R

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Consider f:[a,b] --> R, an integrable and bounded (ain't that implied by "integrable"?!) function and consider {a_n} a sequence that converges towards a and such that a < a_n < b (for all n). Show and rigorously justify that

\lim_{n \rightarrow \infty} \int_{a_n}^{b} f(x)dx = \int_{a}^{b} f(x)dx

All we found is the, imo, not very rigorous and seemingly too easy,

\int_{a}^{b} f(x)dx = \int_{a}^{a_n} f(x)dx + \int_{a_n}^{b} f(x)dx \Rightarrow \lim_{n \rightarrow \infty} \int_{a}^{b} f(x)dx = \lim_{n \rightarrow \infty} \int_{a}^{a_n} f(x)dx + \lim_{n \rightarrow \infty} \int_{a_n}^{b} f(x)dx
\Rightarrow \int_{a}^{b} f(x)dx = 0 + \lim_{n \rightarrow \infty}\int_{a_n}^{b} f(x)dx qed

Does anyone with more insight see how to do this more rigorously or is this the way?
 
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quasar987 said:
Consider f:[a,b] --> R, an integrable and bounded (ain't that implied by "integrable"?!)

Yes. If it's Riemann integrable, then it must be bounded. I don't know anything about Lebesgue integrals.

quasar987 said:
Does anyone with more insight see how to do this more rigorously or is this the way?

I'd say, use the second fundamental theorem of calculus to define an anti-derivative of f... call it F(x).

Then use the first fundamental theorem of calculus, to write your integral of f(x) from a_n to b, as F(b)-F(an). Then the rest should be easy using the properties of limits.
 
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I think you're assuming that there exists a primitive. "Integrable" does not imply "there exist a primitive to f(x) such that the integral is F(b) - F(a)". Maybe f is discontinuous.
 
quasar987 said:
I think you're assuming that there exists a primitive. "Integrable" does not imply "there exist a primitive to f(x) such that the integral is F(b) - F(a)". Maybe f is discontinuous.

Yes, you're right. Forget about the first fundamental theorem of calculus.

Your proof is right... except you need to show that:

lim_{n \rightarrow \infty} \int_{a}^{a_n} f(x)dx =0

According to the second fundamental theorem of calculus, we can define:

F(z)=\int_{a}^{z} f(x)dx

We don't need f(x) to be continuous to do this. So:

F(a_n)=\int_{a}^{a_n} f(x)dx

We can put this in your limit and the limit becomes:

lim_{n \rightarrow \infty} F(a_n)

But according to the second fundamental theorem of calculus F(z) is continuous. So:

lim_{n \rightarrow \infty} F(a_n)= F(lim_{n \rightarrow \infty}a_n)

which equals:

F(a)

and F(a)=\int_{a}^{a} f(x)dx =0

So that should do it.
 
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Very nice! Thank you for that.
 

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