# A density proof of the Dominated Convergence theorem

Consider the following corollary of the DCT, sometimes called the Bounded Convergence theorem:

If $f_n$ is a uniformly bounded sequence of functions converging pointwise a.e. to $f$ on a finite measure space, then $f_n$ converges to $f$ in $L^1$.

If we assume that theorem, can we prove the full DCT like this?

Theorem. Let $f_n\to f$ pointwise a.e. on a measure space $(\Omega, \mu)$, and suppose $|f_n|\leq F$ a.e. for all $n$, where $F$ is some integrable positive function. Then $f_n\to f$ in $L^1$.

Proof. Since $F$ is positive and integrable, we may use it as a density to obtain a finite measure $\mu_F$. The functions $g_n:=\frac {f_n} F$ are well defined $\mu_F$ a.e. and are uniformly bounded by $1$. Therefore by the Bounded Convergence theorem, $g_n\to g:=\frac f F$ in $L^1(\mu_F)$. But $\int |g_n-g| d\mu_F=\int |f_n-f|d\mu$, thus $f_n\to f$ in $L^1(\mu)$. $\blacksquare$

• Looks solid to me. – spaceisdarkgreen Nov 25 '17 at 20:58
• Don't you need $g_n$ to be uniformly bounded everywhere and not a.e.?. What about the points where $|f_n| > F$ – David Reed Nov 25 '17 at 21:16
• @DavidReed Interesting. According to Wikipedia, a.e. domination is only enough to apply DCT if the underlying measure is complete. I didn't realize. But that just means my statement of DCT was too strong - if we strengthen $|f_n|\leq F$ a.e. to $|f_n|\leq F$ everywhere, we get the standard statement of DCT and then $|g_n|\leq 1$ set-theoretically everywhere. – Jack M Nov 25 '17 at 21:25
• Nice job man. I love simple proofs like this. – David Reed Nov 25 '17 at 21:28