# A question in Riemann Integral

Let f be a bounded function on $[a,b]$. Supposed we know that there exists sequences of $U_n$ and $L_n$ of Darboux sums such that $\lim (U_n - L_n) = 0$. How can we show that:

1) f is integrable

2) $\int_{a}^{b} f = \lim U_n = \lim L_n$

Here is what I've done so far:

since $\lim (U_n - L_n) = 0 \rightarrow \forall\epsilon > 0, \exists N \text{ s.t for} n > N |U_n - L_n| < \epsilon$. Since we know that $L_n < U_n$ then we can see that $|U_n - L_n| = U_n - L_n < \epsilon$ and this by definition means that f is integrable.

I'm not sure my argument above is correct or not.

• There needs to be some sentence where you control $U_n$, for instance, since $\lim_n (U_n - L_n) = 0$ could allow $U_n \rightarrow \infty$, as in the case of $f(x) = 1/x$ on $[0,1]$. On the other hand, your $f$ is bounded, so this should not be a problem. – A Blumenthal Mar 7 '13 at 16:15
• @ABlumenthal Every (Riemann) integrable function is bounded. – Did Mar 7 '13 at 16:56

By definition, $f$ is integrable if $\inf U$ and $\sup L$ are equal. Your argument doesn't discuss this. The key here is to use the inequality:
$$L_n \le \sup L \le \inf U \le U_n$$
Since $U_n - L_n \to 0$ as $n \to \infty$, we have $\lim L_n = \lim U_n$ ($f$ is bounded so both limits are finite). Therefore, $\sup L$ and $\inf U$ must be equal to the same limit by the squeeze theorem.