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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!

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    $\begingroup$ You ask "if the improper integral exists, isn't it equal to the proper integral?". Counterexample is the convergent improper integral $\int_0^1 \frac{dx}{\sqrt{x}}$. The integrand is unbounded and not Riemann integrable. $\endgroup$
    – RRL
    Mar 24, 2021 at 16:07

2 Answers 2

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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 $[a+1/n,b]$

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  • $\begingroup$ I do have one question. So in the problem, we are trying to prove that the Riemann and Lebesgue integrals are equal in the case of the Riemann integral being improper. However, you said you used the fact that they are equal. How can we do this if we haven't proven that they are equal in the improper case? $\endgroup$
    – Nolan P
    Mar 24, 2021 at 19:30
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    $\begingroup$ You are given that $f$ is Riemann integrable on every subinterval $[a+\epsilon,b]$ where $\epsilon > 0$. In that case the Riemann integral over that subinterval equals the Lebesgue integral. I use that in my proof where I assert that $\int_{[a,b]} f \chi_{[a+1/n,b]} =\int_{a+ 1/n}^b f(x) \, dx$. This is not to say that $f$ is Riemann integrable on $[a,b]$. We are only given that $\lim_{\epsilon \to 0+} \int_{a+\epsilon}^b f(x) \, dx$ exists -- meaning the improper integral exists. I show that in this case the Lebesgue integral over $[a,b]$ must also be finite. $\endgroup$
    – RRL
    Mar 24, 2021 at 19:41
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    $\begingroup$ So because $f$ is Riemann integrable over each subinterval, it is equal to the Lebesgue integral over each subinterval. I thought you were saying it was over $[a,b]$, that is why I got confused. That clarified it. Thanks! $\endgroup$
    – Nolan P
    Mar 24, 2021 at 19:43
  • $\begingroup$ @DominicBlanco: You're welcome. I was about to respond but you got it! $\endgroup$
    – RRL
    Mar 24, 2021 at 19:44
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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.)

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  • $\begingroup$ We have learned up to Monotone Convergence Theorem. So we know that, BND convergence theorem, Fatou's Lemma, and stuff before that. Basically we've learned to integrate non-negative functions. Haven't extended it in general yet. Most books I've looked at seem to follow that general order, but if you need a specific place, Stein and Shakarchi page 64 is where we are at. $\endgroup$
    – Nolan P
    Mar 24, 2021 at 17:36

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