# About the "Bounded Convergence Theorem"

• The Bounded Convergence Theorem" states that "If a sequence $$\{f_n\}$$ of measurable functions is uniformly bounded and if $$f_n \rightarrow f$$ in measure then $$lim_{n \rightarrow \infty } \int f_n dP = \int f dP$$"

Here does the phrase uniformly bounded" mean that there exists $$M$$ s.t $$\forall n, \vert f_n \vert \leq M$$ ?

• I guess examples like $$f_n = n^{1/2}1_{[0,1/n]}$$ show that being uniformly bounded is not a necessary condition.

• If a sequence of functions $$f_n$$ were to converge uniformly to $$f$$ then one would have had $$lim_{n \rightarrow \infty } \int f_n dP = \int f dP$$. Is the "Bounded Convergence Theorem" a generalization of this?

As in are there examples of sequences of functions not converging uniformly and yet it is true that $$lim_{n \rightarrow \infty } \int f_n dP = \int f dP$$ because they were uniformly bounded meaurable functions converging in measure ?

• For example, the result holds if $f_n\ge 0$ and $f_n\nearrow f$ (equivalently, if conditions of the Monotone convergence Theorem hold). Nov 8 '15 at 22:53
• Can you kindly explain a bit more? Nov 8 '15 at 22:57
• I mean that you can construct an example where $f_n$ converges to $f$ (in pointwise sense and not uniformly) and still get the result. Nov 8 '15 at 22:59
• Any example you have in mind? (So you are saying that uniform convergence is not a necessary condition) But is your example then satisfying the "bounded convergence theorem" conditions? Nov 8 '15 at 23:09
• $f_n$ need not be uniformly bounded... For example, $f≥0$ is such that $\int f =\infty$. Take $f_n=f1\{f≤n\}$. Nov 8 '15 at 23:34

The assumption of the statement is that $f_n$ and $f$ are point-wise bounded by some function $g$ and that $g$ is integrable. You will find more hits if you look for "dominated convergence". See https://en.wikipedia.org/wiki/Dominated_convergence_theorem
For instance, if $g\in L^1$, then $f_n(x)=\min(1,|x|/n)·g(x)$ is uniformly bounded by $g$, converges pointwise to the zero function, but does not necessarily converge uniformly (in the supremums norm).