# Monotone Convergence theorem for decreasing sequence

Suppose $f_n: X\to [0, \infty]$ is measurable for $n = 1, 2, 3, \dots$, $f_1 \geqslant f_2 \geqslant f_3 \geqslant \dots \geqslant 0,$ $f_n(x) \to f(x)$ as $n\to \infty$, for every $x\in X$, and $f_1 \in L^1(\mu)$. Prove that then $$\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu= \int \limits_{X}fd\mu$$ and show that this conclusion does not follow if the condition "$f_1 \in L_1 (\mu)$" is omitted.

Proof: $$f_1 \geqslant f_2 \geqslant f_3 \geqslant \dots \geqslant 0 \implies -f_1 \leqslant -f_2 \leqslant -f_3 \leqslant \dots \leqslant 0 \implies 0\leqslant f_1-f_2\leqslant f_1-f_3\leqslant \dots\leqslant f_1.$$ In other words, sequence $g_n=f_1-f_n$ is increasing & measurable and $g_n(x)\to f_1(x)-f(x)$ for each $x\in X$ and we can use Monotone convergence theorem: $$\lim \limits_{n\to \infty}\int \limits_{X}g_nd\mu=\lim \limits_{n\to \infty}\int \limits_{X}(f_1-f_n)d\mu=\int \limits_{X}(f_1-f)d\mu.$$ If $f_1\in L^1(\mu)$ then $f_n\in L^1(\mu)$ for each $n\in \mathbb{N}$ and $f\in L^1(\mu)$ then: $$\int \limits_{X}f_1d\mu-\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu=\int \limits_{X}f_1d\mu-\int \limits_{X}fd\mu$$ since $\int \limits_{X}f_1d\mu$ is finite we can subtract it and we get what we need!

$\color{red}{Wrong \quad Counterexample:}$ Condition $f_1\in L^1(\mu)$ is crucial! Suppose the we omit this condition. Let $X=\mathbb{N}, \mathfrak{M}=2^{\mathbb{N}}$ and $\mu=|\cdot|$ is counting measure on $\mathfrak{M}$. Suppose $f_n(x)=\dfrac{1_{A_n}(x)}{n}$ where $A_n=\{1,2,\dots, n\}$. It's easy to check that $f_n(x)\to 0$ as $n\to \infty$ for $x\in X$. But $\int \limits_{X}f_nd\mu=\frac{1}{n}\mu(A_n)=1.$ So $$1=\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu\neq \int \limits_{X}fd\mu=0$$

Is my proof and its counterexample correct? Would be very grateful for any suggestions & comments.

EDIT: Let's consider triple $(X,\mathfrak{M},\mu):=(\mathbb{N},2^{\mathbb{N}},|\cdot|)$, where $|\cdot |$ - counting measure on $2^\mathbb{N}$. Let $A_n=\{n, n+1,\dots\}$. Suppose that $f_n(x)=\dfrac{1_{A_n}}{n}$. It's easy to see that $f_1\geqslant f_2\geqslant \dots \geqslant f_n\geqslant \dots \geqslant 0$.

Also note that $f_n(x)\to f(x)=0$ as $n\to \infty$ for $x\in X$. Also $\int \limits_{X}f_nd\mu=\dfrac{1}{n}\mu(A_n)=\infty.$ Hence $$\infty=\lim \limits_{n\to \infty}\int \limits_{X}f_nd\mu\neq\int \limits_{X}fd\mu=0.$$

Is it true?

• Your counterexample is wrong. The sequence is not decreasing and actually $f_1$ is $L^1$. The proof is correct, but you might want to expand on why $f$ is $L^1$. Moreover once you have that, you can simply split the integral of $f_1-f$ by linearity. Commented Jun 13, 2016 at 18:46
• @b00nheT, Wow you're right! Didn't note that $f_1$ is a n element of $L^1$.
– RFZ
Commented Jun 13, 2016 at 18:49
• @b00nheT, I added a new counterexample. I guess that it's already true. What do you think?
– RFZ
Commented Jun 14, 2016 at 7:27
• @b00nheT, Thanks for checking my topic!
– RFZ
Commented Jun 14, 2016 at 7:30
• Does this answer your question? Monotone Convergence Theorem for non-negative decreasing sequence of measurable functions Commented Nov 26, 2022 at 11:35

Almost similar counter example is given if we consider $\mathbb R$ with lebesgue measure and

$$f_n=\mathbb 1_{[n, \infty)}$$

Observe that sequence of functions is decreasing and converges to $0$ but doesn’t satisfy the equality.

By the way this reminds us of continuity properties of measure because there also for the decreasing sequence we have finiteness condition on measure.

From the comments above by @b00nheT.

Your proof is correct. You might want to expand on why $f \in L^1(\mu)$, though.

The first counterexample you gave is incorrect because

1. $f_1$ is in fact in $L^1(\mu)$. We have $$\int_{X} f_1 \mathrm{d}\mu = 1.$$
2. $\{ f_n \}$ is not a decreasing sequence. For instance, observe that $f_1(2) = 0$ and $f_2(2) = 1/2$.

The second counterexample you have given works.

The proof can be reduced to one line if you simply pick $$g = f_1$$ and use Lebesgue's Dominated Convergence Theorem (the notation I use is borrowed from Rudin).

As far as the counterexample goes, I like $$f_n = \chi_{[n,\infty)} : \mathbb R \to [0,\infty]$$.