Monotone Convergence theorem for decreasing sequence

Question

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?

Answer 1

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.

Answer 2

From the comments above by @b00nheT.

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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.

Answer 3

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 $f_n = \chi_{[n,\infty)} : \mathbb R \to [0,\infty]$.