Exercise: Let $p\in[1,\infty)$ and $\lambda$ be the Lebesgue measure on $\mathbb{R}$. Determine for which $\alpha\in\mathbb{R}$ one has $f\in L^p(\mathbb{R})$, where $f(x) = \mathbb{1}_{(0,1)}(x)x^\alpha$.

The given solution: By the Monotone Convergence Theorem we have: $$\int_\mathbb{R}\left|f\right|^pd\lambda = \int_\mathbb{R}\mathbb{1}_{(0,1)}\left|f\right|^pd\lambda = \lim_{n\to\infty}\int_\mathbb{R}\mathbb{1}_{(1/n,1)}\left|f\right|^pd\lambda$$ Now first assume that $\alpha p \neq -1$. Then for each $n\geq 1$ $$I_n:=\int_\mathbb{R}\mathbb{1}_{(1/n,1)}\left|f\right|^pd\lambda = \int_{1/n}^1x^{\alpha p}dx = \dfrac{1}{\alpha p + 1}(1 - n^{-\alpha p - 1})$$ It follows that $\lim\limits_{n\to\infty} I_n$ is finite if and only if $\alpha p + 1 > 0.$ The latter is equivalent with $\alpha > -1/p$. Hence $f\in L^p(\mathbb{R})$ iff $\alpha > -1/p.$

Question: I understand all the derivations in this solutions and I understand the choice for which $\alpha$ gives that $f\in L^p(\mathbb{R})$. What I don't understand is why you have to use the monotone convergence theorem in order to calculate the integral. Why aren't you allowed to say: $$\int_\mathbb{R} \mathbb{1}_{(0,1)}(x)x^{\alpha p}d\lambda = \int_0^1x^{\alpha p}dx$$ Or more generally, when is it allowed?

(It's been a while since I had Riemann integration so I assume that I have to revisit that theory again).

  • $\begingroup$ I think they are being careful about the improper Riemann integral. $\endgroup$ Mar 6, 2018 at 12:28
  • $\begingroup$ The answer to the question in the title is "Always". The answer to the question in the body is different. Hence the title and the body are asking different questions. Not good. $\endgroup$
    – Did
    Mar 9, 2018 at 7:13

1 Answer 1


The monotone convergence theorem is used because it is a way to actually compute the Lebesgue integral using a Riemann integral. The author of this proof is probably adhering to conventions used in the statement of the preceding theorems and is generalizing to an integral over $\mathbb{R}$. Some, but not all, of this is necessary.

Take the case where $p=1$ and $-1 < \alpha < 0$.

We have $f(x) = \mathbf{1}_{(0,1)}(x)x^\alpha$ defined (finite) on any interval $[1/n,1]$. We could just as well have used $\mathbf{1}_{(0,1]}$ since $x^{\alpha}$ is finite at $x = 1$ and the value at one point will not change the integral. In fact, it would have been better, as you will see, since an additional step is necessary if we write the proof carefully.

Note that $f$ is Riemann integrable on $[1/n,1]$ since it is continuous almost everywhere. When the Riemann integral exists, then so does the Lebesgue integral and the values are the same. Hence, we can now work with

$$\underbrace{\int_{\mathbb{R}}\mathbf{1}_{[1/n,1)} |f| d \lambda}_{\text{Lebesgue}} = \underbrace{\int_{\mathbb{[1/n,1)}} |f| d \lambda}_{\text{Lebesgue}} = \underbrace{\int_{\mathbb{[1/n,1]}} |f| d \lambda}_{\text{Lebesgue}} = \underbrace{\int_{1/n}^1 x^{\alpha} \, dx}_{\text{Riemann}} = \frac{1}{\alpha +1}(1 - (1/n)^{\alpha +1}), $$

where we have evaluated the Riemann integral using the fundamental theorem of calculus.

The monotone convergence theorem can now be applied to find

$$\int_{\mathbb{R}}\mathbf{1}_{(0,1)} |f| d \lambda = \int_{\mathbb{R}}\lim_{n \to \infty}\mathbf{1}_{[1/n,1)} |f| d \lambda = \lim_{n \to \infty}\int_{\mathbb{R}}\mathbf{1}_{[1/n,1)} |f| d \lambda \\ = \lim_{n \to \infty}\frac{1}{\alpha +1}(1 - (1/n)^{\alpha +1}) \\ = \frac{1}{\alpha +1}$$

Technically, you can't write $\int_{\mathbb{R}}\mathbf{1}_{(0,1)} |f| d \lambda = \int_0^1 x^\alpha \, dx$ if, by using this notation, the integral on the RHS is interpreted as a Riemann integral. In this case, the Riemann integral does not exist since the integrand is unbounded at $x = 0$. Of course, if it is understood that the RHS integral is an improper integral then this would be acceptable.

To be consistent with notation for Lebesgue integrals, it is preferable to write

$$\int_{\mathbb{R}}\mathbf{1}_{(0,1)} |f| d \lambda = \int_{(0,1)} |f| \, d\lambda$$


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