# Equivalence of a Lebesgue Integrable function

I have the following question:

Let $X$: $\mu(X)<\infty$, and let $f \geq 0$ on $X$. Prove that $f$ is Lebesgue integrable on $X$ if and only if $\sum_{n=0}^{\infty}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) < \infty$.

I have the following ideas, but am a little unsure. For the forward direction:

By our hypothesis, we are taking $f$ to be Lebesgue integrable. Assume $\sum_{n=0}^{\infty}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) = \infty$. Then for any n, no matter how large, $\mu(\lbrace x \in X : f(x) \geq 2^n \rbrace)$ has positive measure. Otherwise, the sum will terminate for a certain $N$, giving us $\sum_{n=0}^{N}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) < \infty$. Thus we have $f$ unbounded on a set of positive measure, which in combination with $f(x) \geq 0$, gives us that $\int_E f(x) d\mu=\infty$. This is a contradiction to $f$ being Lebesgue integrable. So our summation must be finite.

For the reverse direction:

We have that $\sum_{n=0}^{N}2^n \mu(\lbrace x \in X : f(x) \geq 2^n \rbrace) < \infty \\$. Assume that $f$ is not Lebesgue integrable, then we have $\int_E f(x) d\mu=\infty$. Since we are integrating over a finite set $X$, then this means that $f(x)$ must be unbounded on a set of positive measure, which makes our summation infinite, a contradiction.

Any thoughts as to the validity of my proof? I feel as if there is an easier, direct way to do it.

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Regarding the first implication of your proof: I think this works. But I might be missing something. –  Matt N. Aug 14 '12 at 19:51
Sorry but the proofs of both implications are flawed. Forward: integrable does not imply bounded. Reverse: not integrable does not imply infinite on a set of positive measure. –  Did Aug 14 '12 at 21:46
To be precise, the first proof offered only wanted $\textit{integrable}$ to imply $\textit{essentially bounded}$. Of course the implication still fails. –  Kevin Carlson Aug 15 '12 at 0:31

The fact that the summation is infinite does not mean that it does not converge, consider $f(x)=\frac{1}{\sqrt x}$ on $[0,1]$. It is integrable, yet clearly the summation is not finite in this case: it is $\sum_n 2^n\cdot 2^{-2n}=2<\infty$.

Put $A_n:=\{ x\vert f(x)\geq 2^n\},B_n:=A_n\setminus A_{n+1},a_n=\mu(A_n)$. Then define $h_1=\sum_n2^{n+1}\chi_{B_n}$. Except for the points where $f<1$ or $f=\infty$ (which are resolved easily), we have $h_1/2\leq f\leq h_1$, so if $f$ is not infinite on a set of positive measure, $f$ is integrable iff $h_1$ is.

On the other hand, $$\int f\leq \int h_1=2\sum_{n\geq 0} 2^n(a_n-a_{n+1})\leq 2\sum_n 2^n a_n$$

Edit2: Instead, take a look at $h_2:=\sum_n 2^{n-1}\chi_{A_n}$. Then for each $x\in B_n$ we have that $h_2(x)\leq \sum_{j\leq n}2^{j-1}<2^n\leq f(x)$, and $h_2\leq f$ almost everywhere, so if $f$ is integrable, $h_2$ is too. Then: $$\int h_2=\sum_n 2^{n-1}\mu(A_n)=\frac{1}{2}\sum_n 2^n\mu(A_n)$$ so $\sum_n 2^n\mu(A_n)$ converges and we're done.

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Excellent! This really helped me understand the problem quite well. Thank you! –  Frank White Aug 14 '12 at 20:35
Yeah, I was just noticing that. Dang. –  Frank White Aug 14 '12 at 21:11
@AllenCox: I think I managed to circumvent the problem. :) –  tomasz Aug 14 '12 at 21:32
I'm a little confused on how the intersections of $A_n$ having zero measure gives us $h_2 \leq f$ a.e. Don't we get that anyway based on how $h_2$ is defined? –  Frank White Aug 14 '12 at 21:46
@AllenCox: right, I overthought that part. :) in the intersection both $h_2$ and $f$ are infinite, so it's okay. –  tomasz Aug 14 '12 at 21:49

$$\frac12\left(1+\sum_{n=0}^{+\infty}2^n\,\mathbf 1_{f\geqslant2^n}\right)\leqslant f\lt1+\sum_{n=0}^{+\infty}2^n\,\mathbf 1_{f\geqslant2^n}$$

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Excellent! Why didn't I think of that? –  Byron Schmuland Aug 14 '12 at 21:43
@Byron: Thanks. Dunno... :-) –  Did Aug 14 '12 at 21:47

Hint: Switching the order of integration gives $$\int_X f(x)\,\mu(dx)=\int_X \int_0^{f(x)}\,dt\,\mu(dx) =\int_0^\infty \mu(x: f(x)>t)\,dt$$ This equation is true whether the two sides are finite or infinite.

Since $G(t):= \mu(x: f(x)>t)$ is decreasing, we have $$G(2^{n+1}) \int_{2^n}^{2^{n+1}}\,dt \leq \int_{2^n}^{2^{n+1}} G(t) \,dt \leq G(2^n) \int_{2^n}^{2^{n+1}}\,dt$$ and hence $\int_0^\infty G(t)\,dt<\infty$ if and only if $\sum_{n=0}^\infty 2^n G(2^n)<\infty.$

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This won't quite do. Your argument doesn't actually use the $2^n$ at all: it would read exactly the same if we asked only for $\sum_{n=0}^{\infty}\mu(\{x \in X: f(x) \geq 2^n\}) < \infty$. Yet $\frac{1}{x}$ satisfies this weaker hypothesis on $[0,1]$: the infinite sum becomes $\sum_{i=0}^\infty \frac{1}{2^i}=2$, while we know the integral diverges. See if you can try a similar approach that explicitly uses the exponential decay of the measure of sets on which $f$ is big.

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What do you mean by using the exponential decay on the measure of the sets? Should the sets have some sort of known bound? –  Frank White Aug 14 '12 at 18:29
Just that since the sum of products of an exponentially growing quantity with the measures of the sets above $2^n$ was finite, we'd better have the latter measures shrink faster than $2^{-n}$, which I referred to as "exponential(ly)". –  Kevin Carlson Aug 15 '12 at 0:29

Let $E_n = \lbrace x \in X : 2^n \leq f(x) < 2^{n+1} \rbrace$. Let $F = \lbrace x \in X : f(x) < 1 \rbrace$. Then $X$ is a disjoint union of $F$ and $E_n$, $n = 0,1,\dots$. Hence $\int_X f(x) d\mu = \int_F f(x) d\mu + \sum_{n=0}^{\infty} \int_{E_n} f(x) d\mu$.

Let $G_n = \lbrace x \in X : f(x) \geq 2^n \rbrace$. Then $E_n = G_n - G_{n+1}$. Let $m_n = \mu(G_n)$. Since $\mu(E_n) = \mu(G_n - G_{n+1}) = m_n - m_{n+1}$, $2^n(m_n - m_{n+1}) \leq \int_{E_n} f(x) d\mu \leq 2^{n+1}(m_n - m_{n+1})$. Hence

$\int_F f(x) d\mu + \sum_{n=0}^{\infty} 2^n(m_n - m_{n+1}) \leq \int_X f(x) d\mu \leq \int_F f(x) d\mu + \sum_{n=0}^{\infty} 2^{n+1}(m_n - m_{n+1})$

Let $s = \sum_{n=0}^{\infty} 2^n m_n$. Suppose $f$ is integrable. By the above left inequality, $\sum_{n=0}^{\infty} 2^n(m_n - m_{n+1}) < \infty$. Since $\sum_{n=0}^{\infty} 2^n(m_n - m_{n+1}) = m_0 + \sum_{n=0}^{\infty} 2^n m_{n+1} = m_0 + (\sum_{n=0}^{\infty} 2^{n+1} m_{n+1})/2 < \infty$, $s < \infty$.

Conversely suppose $s< \infty$. Note that $\int_F f(x) d\mu < \infty$. By the above right inequality, $\int_X f(x) d\mu \leq \int_F f(x) d\mu + \sum_{n=0}^{\infty} 2^{n+1}(m_n - m_{n+1}) = \int_F f(x) d\mu + 2m_0 + \sum_{n=0}^{\infty} 2^{n+1} m_{n+1} = \int_F f(x) d\mu + m_0 + s < \infty$.

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I do not understand how $\sum_{n=0}^{\infty}2^nA=2A<\infty$. This seems to be an error. –  Frank White Aug 14 '12 at 18:54
You are right. I will edit my answer using the idea of the latter part of my answer. –  Makoto Kato Aug 14 '12 at 19:04
What do you think of my proof? –  Makoto Kato Aug 15 '12 at 19:06
Well, it's essentially the same proof as tomasz's but with different notation, is it not? –  Frank White Aug 15 '12 at 19:35
Actually I haven't read his proof. I posted an answer quite early, you know. There was a stupid error as you pointed out, but the basic idea was the same as the current one. –  Makoto Kato Aug 15 '12 at 20:00
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