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Theorem (Bounded Convergence Theorem) Let $\{f_n\}$ be a sequence of measurable functions on a set of finite measure $E$. Suppose $\{f_n\}$ is uniformly pointwise bounded on $E$, that is , there is a number $M\geq 0$ for which $|f_n| \leq M$ for all $n$. If $\{f_n\} \to f$ pointwise on $E$, then $\lim\limits_{n \to \infty} \int_E f_n = \int_E f.$

Why is it important in this theorem for it to be uniformly pointwise bounded as opposed to just pointwise bounded? Is this because if it is not uniformly bounded than it is certainly not uniformly convergent.

Norbert
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emka
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  • What do you mean by "just bounded"? There are two ways to relax the condition of uniformly bounded: (1) require that each $f_n$ be a bounded function, or (2) require that ${f_n}$ is pointwise-bounded, meaning that $|f_n(x)| \leq \phi(x)$ for some $\phi$. – Jesse Madnick Nov 12 '12 at 07:35
  • Also, uniform boundedness does not imply uniform convergence. – Jesse Madnick Nov 12 '12 at 07:36
  • @Jesse I mean pointwise bounded. Uniform convergence implies uniform boundedness. So not uniformly bounded implies not uniformly convergence. – emka Nov 12 '12 at 07:46
  • Uniform convergence implies uniform boundedness assuming each $f_n$ is a bounded function. For example, consider the sequence $f_n\colon (0,1) \to \mathbb{R}$ given by $f_n(x) = 1/x + 1/n$, which is uniformly convergent but not uniformly bounded. – Jesse Madnick Nov 12 '12 at 18:29
  • Can we say also $\lim_{n\to \infty}\int_Ef^p_n=\int_Ef^p$ for $p>1$ as well? – Saj_Eda Sep 14 '18 at 21:06

2 Answers2

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If you avoid the requirement of uniform boundedness then there is a counterexample $$ f_n=n^2 1_{[0,n^{-1}]} $$

But there are examples when the theorem holds even if the sequence of functions is not uniformly pointwise bounded. For example $$ f_n=n^{1/2}1_{[1,n^{-1}]} $$

The most general requirement on boundedness of $f_n$ when theorem still holds is $$ \forall n\in\mathbb{N}\quad\forall x\in E\quad |f_n(x)|\leq F(x) $$ for some integrable $F:E\to\mathbb{R}_+$. You can also weaken the condition of pointwise convergence just to convergence in measure $$ \forall\varepsilon>0\quad\lim\limits_{n\to\infty}\mu(\{x\in E:|f_n(x)-f(x)|>\varepsilon\})=0 $$

Norbert
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Consider the example $$f_n(x)=\cases{ nx & if $0\leq x \leq {1\over n}$\cr -n\Bigl(x-{2\over n}\Bigr) & if ${1\over n}\leq x \leq {2\over n}$\cr 0& if ${2\over n}\leq x\leq 1$}$$ Each $f_n$ is piecewise linear and continuous on the interval $[0,1]$, and $f_n(x)\to 0$ as $n\to\infty$ for each $x\in[0,1]$. However, $\int_0^1 f_n(x){\rm d}x=1$ for each $n$, so the conclusion of the Bounded Convergence Theorem does not hold here.

I think this example shows quite clearly what the problem is. You do not want to assume that the convergence is uniform, so there might be a substantial difference between $f(x)$ and each $f_n(x)$ on some set, and you need to make sure that the integral of $|f_n-f|$ over this set will be small as $n\to\infty$.

(Usually this theorem is stated in a slightly stronger form, called the Dominated Convergence Theorem, where the uniform bound $|f_n|\leq M$ is replaced by $|f_n(x)|\leq g(x)$ for each $x$ and $n$, where $g$ is an integrable function.)

Per Erik Manne
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  • The sequence of functions you have considered is not even pointwise bounded. Is it possible to say that if the sequence of functions is pointwise bounded in stead of uniformly bounded, then also bounded convergence theorem is true? Per Manne – Anupam Nov 03 '16 at 09:49
  • @Anupam I think you may have misunderstood the definition of pointwise bounded. For each $x$, the sequence ${f_n(x)}$ has only finitely many nonzero terms, so it is surely bounded. – Per Erik Manne Nov 03 '16 at 21:02
  • Menne: What I understand by pointwise bounded is as follows: A sequence of functions $(f_n)$ is called pointwise bounded if for each $x$ there exists $M_x>0$ such that $|f_n(x)|\leq M_x$ for all $n\in \mathbb N$. Is it the same thing you are also saying? – Anupam Nov 04 '16 at 06:02
  • Menne: Yes I understood where I have mistaken. Thanks! – Anupam Nov 04 '16 at 06:53