Let $f$ and $f_n$ measurable numeric functions, $n\in \mathbb N$ and $f=\lim_{n\to\infty}f_n$ a.e. and suppose an integrable $g\ge 0 $ exists with $\mid f_n \mid \le g$ a.e. Then $f$ and $f_n$ are integrable with $\lim_{n\to\infty}\int \mid f_n -f \mid d\mu=0$.

i) Why is this equivalent to $\int fd\mu=\lim_{n\to\infty }\int f_n d\mu$?

ii) And why can one suppose in the proof of this statement (DMCT) that $f$ and $f_n$ are real valued?

  • $\begingroup$ What does "DMCT" stand for here? People sometimes write "DCT" for "dominated convergence theorem" but the M has me stumped... $\endgroup$ – David C. Ullrich Aug 13 '18 at 14:39
  • $\begingroup$ yes I mean Lebesgue's dominated convergence theorem $\endgroup$ – user563311 Aug 13 '18 at 15:41
  • $\begingroup$ Right. So why is it "DMCT"? What does the M stand for? $\endgroup$ – David C. Ullrich Aug 13 '18 at 15:44
  • $\begingroup$ I think it was a typo. Actaully I mean DCT $\endgroup$ – user563311 Aug 13 '18 at 16:32

Saying $\lim\int|f-f_n|=0$ is not equivalent to saying $\lim\int f_n=\int f$ in general. The first implies the second (since $|\int f_n-\int f|=|\int(f_n-f)|\le\int|f_n-f|$) but not conversely.

Wondering where you could have got the idea that they were equivalent, I think I got it. The following two theorems are equivalent:

Thm 1. Suppose $f_n\to f$ almost everywhere, $g\ge0$, $\int g<\infty$, and $|f_n|\le g$. Then $\int f_n\to\int f$.

Thm 2. Suppose $f_n\to f$ almost everywhere, $g\ge0$, $\int g<\infty$, and $|f_n|\le g$. Then $\int|f_n-f|\to0$.

Note that saying the theorems are equivalent does not say that the conclusions of the theorems are equivalent.

It's trivial that Theorem 2 implies Theorem 1. Since I suspect that showing the two theorems are equivalent was a homework problem, I'll just give a huge hint how to show that Theorem 1 implies Theorem 2:

Suppose Theorem 1. Suppose $f_n\to f$ almost everywhere, $g\ge0$, $|f_n|\le g$, and $\int g<\infty$. Let $F_n=???$ and let $G=???$. Then $F_n\to F$ almost everywhere, where $F=???$. Also $G\ge0$, $|F_n|\le G$, and $\int G<\infty$. So Theorem 1 implies that $\int F_n\to\int F$, and it follows that $\int|f_n-f|\to0$.

(That does not show that $\int f_n\to\int f$ and $\int|f_n-f|\to0$ are equivalent, because in deriving Theorem 2 from Theorem 1 we applied Theorem 1 to the sequence $(F_n)$, not to the seqence $(f_n)$.)

  • $\begingroup$ I encountered this problem when I was working on DMCT, sometimes the statement only from Thm 1. is stated and sometimes both Thm 1. and Thm 2., but thanks for your answer anyway $\endgroup$ – user563311 Aug 13 '18 at 15:40


First, note that

$$\left| \int f_n - \int f \right|\leq \int \left|f_n-f \right|$$


$$\lim_{n\rightarrow \infty}\int |f_n-f|=0\implies \lim_{n\rightarrow \infty}\int f_n = \int f $$

Conversely, note that

$$|f_n-f|\leq 2g $$

and that

$$\lim_{n\rightarrow \infty} |f_n-f|=0$$

Then by DMCT,

$$\lim_{n\rightarrow \infty}\int|f_n-f| = \int 0 = 0 .$$

2) Use that

$$\int f = \int Re(f) + i\int Im(f)$$

and that

$$\lim_{n\rightarrow \infty} z_n= \lim_{n\rightarrow \infty}Re(z_n)+i\lim_{n\rightarrow \infty}Im(z_n).$$


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