# Proof of dominated convergence theorem

I was going through the proof of the Dominated Convergence Theorem.

Now if we have that ($$f_n$$) is a sequence of measurable functions such that $$\lvert f_n\rvert$$ $$\le$$ $$g$$ for all n where g is integrable on $$\Bbb{R}$$. And if $$f$$ = $$\lim_{n}f_n$$ almost everwhere.

We can show that ($$g+f_n$$) is a sequence of non-negative measurable functions.

Then by Fatou's lemma, we have that $$\int$$liminf($$g+f_n$$)$$dx$$ $$\le$$ liminf$$\int$$($$g+f_n$$)$$dx$$.

Now from here, we can obtain that

$$\int$$($$g+f$$)$$dx$$ $$\le$$ $$\intgdx$$+liminf $$\intf_ndx$$.How?

Please explain this last step.I know that since both are integrable, the integral can be seperated..but how is liminf seperated in the right-hand side?

Inside the liminf, $$\int g$$ is just a number. So what it says is that $$\liminf_n(c+f_n)=c+\liminf _nf_n,$$ where $$c=\int g$$.

• So..is that always true? ..liminf(c+an) = c+liminfan? Jun 10 '20 at 5:19
• Yes. How could translating a set by a fixed distance could change it in any other way? If you move a set 10 units to the right, the infimum will increase by 10. Jun 10 '20 at 5:21
• Are you sure $\int$g is a number? Jun 10 '20 at 5:39
• for example..what if g(x)=x? Jun 10 '20 at 5:40
• I can see it is independent of n..it's a constant sequence Jun 10 '20 at 5:41