1

Let $(X, \mathcal{M}, \mu )$ be a space of finite measure and $f \colon X \rightarrow \mathbb{C}$. For each $n \in \mathbb{N}$, let $A_n = \{ x \in X \colon |f(x)|>n \}$ and $B_n = \{ x \in X \colon n \leq |f(x)|< n+1\}$. Show that the following statements are equivalent:

(a) $f \in L^{1}(\mu)$
(b) $\sum_{n=1}^{\infty}n \mu(B_n) < \infty$
(c) $\sum_{n=1}^{\infty}{\mu (A_n)} < \infty$

Definition: We define $L^1(\mu)$ as the collection of all measurable complex functions $f$ in $X$ for which $\int_{X}{|f|}d \mu < \infty$. The elements of $L^1 (\mu)$ are called Lebesgue integrable functions (with respect to $\mu$) or summable functions.

Definition 2: If $f = u + iv$, where $u$ and $v$ are real functions measurable in $X$, and if $f \in L^1(\mu)$, we define $\int_{E}{f} d\mu = \int_{E}{u^{+}} d\mu - \int_{E}{u^{-}} d\mu + i\int_{E}{v^{+}} d\mu - i\int_{E}{v^{-}} d\mu $ for each measurable set $E$.

Here $u^{+}$ and $u^{-}$ are the positive and negative parts of $u$; $v^{+}$ and $v^{-}$ are obtained analogously from $v$. These four functions are measurable, real, and non-negative.

I wanna do a test for this exercise, but I really don't have a good idea to do it. To show that (a) implies (c), for example, I have thought about defining the function, $f$ let's say: $f(x) = \sum_{n=1}^{\infty}{\chi_{A_n}}(x)$, where $\chi_{A_n}$ is the characteristic function of $A_n$. Then use the following theorem: If $f_n \colon X \rightarrow [0, \infty]$ is measurable, for $n = 1, 2, 3,. . . ,$ and $f(x) = \sum_{n=1}^{\infty}{f_{n}(x)}$ for all $x \in X$, then $\int_{X}{f}d\mu = \sum_{n=1}^{\infty}{\int_{X}{f_{n}d\mu}}$.

To then obtain that $\infty > \int_{X}f d\mu = \sum_{n=1}^{\infty}{\int_{X}{\chi_{A_n}}d\mu} = \sum_{n=1}^{\infty}{\mu (A_n)}$. I'm not sure of my reasoning because the function $f$ is really a complex function. I need some help to do this.

Curious
  • 591
  • 1
  • 5
  • 21
  • https://math.stackexchange.com/a/3827802/121671 – Mittens Apr 28 '21 at 03:57
  • Does this answer your question? Prove that... –  Apr 28 '21 at 04:59
  • @Masacroso Your answer is great, however in my measurement course I'm not yet familiar with Tonelli's theorem. I wanted to use another reasoning. Something similar to the jdods test in https://math.stackexchange.com/questions/1387040/prove-that-f-is-integrable-if-and-only-if-sum-infty-n-1-mu-x-in-x-f which I have not been able to understand. On the other hand, the function $f$ is a complex function and not necessarily positive as I have seen in other statements. – Curious Apr 28 '21 at 05:52
  • @Masacroso How can I get a more simplified proof of this exercise? I don't know if my idea is useful in the statement, but I think it should be the way if I'm not wrong. – Curious Apr 28 '21 at 05:58
  • instead of Tonelli's theorem you can use monotone convergence –  Apr 28 '21 at 17:09
  • @Masacroso We have that if $A_k={x:f(x)\geq k}$ and $B_k={x: |f(x)| \in[k,k+1)}$. The $B_k$ are pair-wise disjoint. We have that $\displaystyle A_n=\bigcup_{k=n}^\infty B_k$. This gives us $\displaystyle \mu(A_n)=\sum_{k=n}^\infty \mu(B_k)$. On the other hand I'm not sure whether to say $f = \sum_{n=1}^{\infty}{\chi_{B_k}f}$, where $\chi_{B_k}$ is the characteristic function of $B_k$. – Curious Apr 28 '21 at 18:57
  • I have really been trying to do this exercise and have been frustrated with not being able to accomplish anything. I don't know how to use the monotone convergence therem here. Also the function $f$ is complex and I think that I couldn't assume that it is nonnegative in general. I could also think of writing $f = u +iv$ in order to use definition 2 that added above, but I don't know if it is a good way. – Curious Apr 28 '21 at 19:00
  • Are you sure your definition for $A_n$ in the problem statement is correct? I think it should have $|f(x)|>n$. – MathematicsStudent1122 Apr 28 '21 at 20:55
  • @AlohaSine yes, indeed the statement to be proved is false as far as some $B_n$ had positive measure. It must be $\ge n$ (or $>n$) in the definition of each $A_n$ –  Apr 28 '21 at 21:11
  • @AlohaSine Thank you very much, I have already edited my statement. – Curious Apr 28 '21 at 21:19
  • @OliverDiaz I have been looking at your suggestion, but it is not totally clear to me and I think there are some differences. Among them, I have noticed that the function there is real and they assume it to be non-negative while in my case it is a function with values ​​in all $\mathbb{C}$. On the other hand, I'm limited to using the monotonic convergence theorem and this is just what I have not been able to do. – Curious Apr 28 '21 at 21:34
  • your function is also real as $|f|$ is non-negative. The definition of $f\in L^1$ means that $\int |f|<\infty$ –  Apr 28 '21 at 22:34
  • @Curious: I wrote a sketch of a proof. Some details are for you to fill in. The equivalence of (b) and (c) relies on changing order of summation in a double series of positive terms with is something usually seen in basic analysis courses: $\sum_n\sum_m a_{nm}=\sum_m\sum_na_{nm}$ where $a_{nm}\geq0$. – Mittens Apr 29 '21 at 00:09

2 Answers2

2

Here is an attempt. (a) $\Rightarrow$ (b). $$\sum_{n \in \mathbb{N}}n\mu(B_n)=\sup_{N \in \mathbb{N}}\sum_{n=1}^Nn\mu(B_n)=\sup_{N \in \mathbb{N}}\sum_{n=1}^Nn\int\mathbb{I}_{B_n}(x)\mu(dx)=$$ $$=\sup_{N \in \mathbb{N}}\int \bigg(\sum_{n=1}^Nn\mathbb{I}_{B_n}(x)\bigg)\mu(dx)=\sup_{N \in \mathbb{N}}\int f_N(x)\mu(dx)$$ we have that $f_N \to g\leq|f| \in \mathcal{L}^1(\mu)$ pointwise and $g$ is positive and measurable so $g \in \mathcal{L}^1(\mu)$ and by DCT $$\sup_{N \in \mathbb{N}}\int f_N(x)\mu(dx)=\int g(x)\mu(dx)<\infty$$

(b) $\Rightarrow$ (c). If $|f(x)|>n$ then $\mathbb{I}_{A_n}(x)=1$ and $n\mathbb{I}_{B_n}(x)\geq n$. If $|f(x)|=n$ then $\mathbb{I}_{A_n}(x)=0$ and $n\mathbb{I}_{B_n}(x)=n$ . If $|f(x)|<n$ then $\mathbb{I}_{A_n}(x)=0$ and $n\mathbb{I}_{B_n}(x)=0$. Therefore $$\mathbb{I}_{A_n}(x)\leq n \mathbb{I}_{B_n}(x)$$ $$\int\mathbb{I}_{A_n}(x)\mu(dx)\leq \int n \mathbb{I}_{B_n}(x)\mu(dx)$$

$$\sum_{n=1}^\infty \mu(A_n)\leq \sum_{n=1}^\infty n\mu(B_n) < \infty$$

(c) $\Leftrightarrow$ (a). Is answered here (as suggested in the comments, use monotone convergence if you do not want to use Tonelli).

Snoop
  • 18,347
  • There are three things I don't understand in your test. The first two are about the notation: when you write $f_N$ are you saying that $f$ depends on that $ N$? On the other hand, the writing $\mu (dx)$ is the same as $d \mu$ in my statement? In this same part, when you swap the integra with the sum, are you using the Fubini / Tonelli theorem? – Curious Apr 29 '21 at 05:15
  • Finally and the most interesting for me is about the convergence that you mention from $f_N$ towards a function $g$. Who is the function $g$ and why $g \leq |f|$? In other words, how is the line justified: "$ f_N \to g \leq | f | \in \mathcal{L}^1 (\mu) $ pointwise and $ g $ is positive and measurable so $ g \in \mathcal{L}^1 (\mu) $? How do I also justify the fact that $ f_N \to g $? – Curious Apr 29 '21 at 05:19
  • (1). $f_N$ is a sequence indexed by $N$. – Snoop Apr 29 '21 at 09:50
  • (2). $\mu(dx)$, $d\mu$ are equivalent notation. The former is used when you make the integration variable explicit for the sake of clarity. (3). When I swap integral and sum I use the linearity of the $\mu$-integral: the sup is still at its place. It's when we want to put it inside an integral that we need convergence theorems. – Snoop Apr 29 '21 at 09:58
  • (4). Sorry for being handwavy on this one, it's a good question. We have that the $f_N$ are functions that are increasing pointwise and are always positive (thus their sup is measurable), but they are also bounded by $|f|$, so their sup is also in $\mathcal{L}^1$ and a.e. $\mathbb{R}$ valued. This should allow the use of dominated convergence and so the sup goes inside the integral: $g$ being positive and in $\mathcal{L}^1$ tells us that its $\mu$-integral is finite. – Snoop Apr 29 '21 at 10:05
  • Now I understand. But I'm not sure about the following: why is there the limit or the sup of $f_N$? That is, why does $ g $ exist such that $ f_N \to g \leq | f | $? – Curious Apr 29 '21 at 20:24
  • The $f_N$ are all positive, increasing and measurable, so their pointwise sup exists in $\bar{\mathbb{R}}$ (it's either a real number or +inf). However they are all bounded by $|f|$, which is in $\mathcal{L}^1(\mu)$, so their sup that I call $g$ is such that $f_N\leq \sup f_N = g \leq |f|$ and this implies that $g \in \mathcal{L}^1$. Leave an upvote if you like the answer! ;) – Snoop Apr 29 '21 at 20:31
  • Thank you. Now I understand. I will ask one last question: is it correct to change the word sup for limit? That is, I can write $ \sum_{n \in \mathbb {N}} n \mu (B_n) = \lim_{N \to \infty} \sum_{n = 1}^Nn \mu(B_n) $ in instead of sup. Also, can I write $ \lim_{N \to \infty} {f_N} = g $? – Curious Apr 29 '21 at 20:42
  • In this case, sup and lim are the same because the sequences are increasing in $N$. – Snoop Apr 29 '21 at 20:47
  • 1
    Thank you so much. I have given you a good upvote – Curious Apr 30 '21 at 04:44
2

  1. From $$\begin{align} n \mu(n<|f|\leq n+1)&\leq \int_{n<|f|\leq n+1}|f|\,d\mu\leq (n+1)\mu(n\leq |f|\leq n+1)\\ &\leq 2n\mu(n\leq |f|\leq n+1)\end{align}$$ one gets that $$\sum_n n\mu(n<|f|\leq n+1)\leq \int |f|\,d\mu\leq 2\sum_n n\mu(n<|f|\leq n+1)$$ So convergence of $\int|f|\,d\mu$ and $\sum_nn\mu(n<|f|\leq n+1)$ are equivalent.

  2. For the other equivalencies, notice that $$\begin{align} \sum^\infty_{n=1}\mu(|f|>n)&=\sum^\infty_{n=1}\sum^\infty_{m=n}\mu(m<|f|\leq m+1)=\sum^\infty_{m=1}\sum^m_{n=1}\mu(m<|f|\leq m+1)\\ &=\sum^\infty_{m=1}m\mu(m<|f|\leq m+1)\end{align}$$ The conclusion follows immediately.

Mittens
  • 46,352
  • Very interesting! When you say that the convergence of $\int |f| d \mu$ and $\sum_{n}{n \mu (n<|f| \leq n+1) }$ are equivalent, are you using the sandwich theorem? – Curious Apr 29 '21 at 04:56
  • I understand. But along the same lines, when you consider the sums of the inequalities, then an integral in all $ X $ appears between the sums. Why does this happen like this? Why doesn't the sum of the integral appear? – Curious Apr 29 '21 at 05:32
  • On the other hand I have noticed that in $ \sum^\infty_{n = 1} m \mu (m <| f | \leq m + 1) $ the expression within the sum doesn't depend on the index $n$ . – Curious Apr 29 '21 at 13:14
  • Now that part is clear to me. The confusion is why in (1) you write $$ \sum_n n \mu (n <| f | \leq n + 1) \leq \int | f | ,d \mu \leq 2 \sum_n n \mu (n <| f | \leq n + 1) $$ instead of $$ \sum_n n \mu (n <| f | \leq n + 1) \leq \sum_n \int_{n <\leq | f | <n + 1} | f | ,d \mu \leq 2 \sum_n n \mu (n <| f | \leq n + 1) $$? – Curious Apr 29 '21 at 15:12
  • Because the sum $\sum^\infty_{n=0}\int_{n<|f|\leq n+1}|f|,d\mu =\int|f|,d\mu$. That was a little quick perhaps. – Mittens Apr 29 '21 at 15:24