Improper Riemann Integral equal to Lebesgue Integral Let $f: [a,b] \to [0,\infty)$ and $f$ is Riemann Integrable on every subinterval $[a + \epsilon,b]$ for $\epsilon > 0$. Suppose that the improper Riemann integral exists. That is
$$I = \lim_{\epsilon \to 0} \int_{a + \epsilon}^{b} f(x) dx < \infty$$
exists. Prove that $f$ is Lebesgue integrable on $[a,b]$ and that $\int_{[a,b]} f(x) dx = I$.
I found another posting of this problem written in the same way, but it used dominated convergence theorem, which I have not covered. The others I saw were written slightly differently and/or weren't making too much sense.
Also, maybe I'm not seeing something clearly, but here's something I thought:
The improper integral I is finite (it exists). If the improper integral exists, isn't it equal to the "proper" integral? That would mean the "normal" proof of showing that every Riemann Integrable function is Lebesgue integrable would apply. However, I have a feeling this is not the way to prove it or else this question wouldn't be asked.
Thanks for the help!
 A: Since $f$ is nonnegative, the Lebesgue integral must exist (but may be infinite).  With the existence of the improper Riemann integral we can show that the Lebesgue integral is finite and $f$ is "Lebesgue integrable" on $[a,b]$.
For all sufficiently large $n$ we have $[a+1/n,b] \subset [a,b]$. Since $f \chi_{[a+1/n,b]} \nearrow f$ as $n \to \infty$, it follows by the monotone convergence theorem that
$$\int_{[a,b]} f = \int_{[a,b]} \lim_{n \to \infty}f \chi_{[a+1/n,b]} =  \lim_{n \to \infty}\int_{[a,b]} f \chi_{[a+1/n,b]} = \lim_{n \to \infty}\int_{a+ 1/n}^b f(x) \, dx = \int_a^bf(x) \, dx < +\infty$$
Here we have applied the equivalence of the Lebesgue and Riemann integrals on the interval $[a+1/n,b]$
A: Consider the Riemann integral
$$  \int_0^1 x^{-2} \,\mathrm{d}x  $$
This is an improper Riemann integral due to the unbounded behaviour at the left endpoint of the interval of integration.  It's value is defined to be (if this limit exists)
$$  \lim_{\varepsilon \rightarrow 0^+} \int_{\varepsilon}^1 x^{-2} \,\mathrm{d}x  \text{.}  $$
Now think of a decreasing sequence $(\varepsilon_i)_{i \in \Bbb{Z}_{>0}}$ where each $\varepsilon_i \in (0, 1]$.  This sequence provides a way to get at that limit:
$$  I_i = \int_{\varepsilon_i}^1 x^{-2} \,\mathrm{d}x  \text{.}  $$
Every $I_i$ is a proper integral.  But just because every proper integral in the sequence $(I_i)_i$ converges does not mean the limit $\lim_{i \rightarrow \infty} I_i$ exists.  Contrast with $\displaystyle J_i = \int_{\varepsilon_i}^1 x^{-1} \,\mathrm{d}x$, for which each $J_i$ is some finite number, but the limit of $J_i$ as $i \rightarrow \infty$ does not exist.
So to make progress, you will use some idea like $|I - I_i| < \delta$, which idea is based on the fact that $I$ exists and is finite and that each $I_i$ exists, is finite, and is approaching the integral over the whole interval.  You might use this to show $f$ is Lebesgue integrable on $(0,1]$, then observe that $[0,1] \smallsetminus (0,1]$ is a set of Lebesgue measure zero, obtaining your goal.  (I'm having to guess at what tools are available to you -- you assert that DCT is unavailable, but you don't say what is available.)
