Dominated Convergence Theorem

• categories: Measure Theory, Real Analysis, Theorem

Let $\{f_n\}$ be a sequence of measurable functions on a measure space $(X,\mathcal{M},\mu )$ such that:

1. $f_n(x)\to f(x)$ pointwise almost everywhere on $X$.
2. There exists an integrable function $g\in L^1(X)$ such that $|f_n(x)|\le g(x)$ for all $n$ and almost every $x\in X$.

Then:

$$\underset{n\to \infty }{\lim }{\int }_X{f}_n(x)d\mu (x)={\int }_X\underset{n\to \infty }{\lim }{f}_n(x)d\mu (x)={\int }_{X}f(x)d\mu (x)$$

Intuition: The theorem allows us to exchange the limit and the integral if the functions $f_n$ are uniformly dominated by an integrable function $g$. The domination ensures that the sequence doesn’t “blow up” despite pointwise convergence.

Key Conditions:

1. Pointwise convergence: $f_n\to f$ almost everywhere.
2. Uniform domination: There exists an integrable bound $g$ such that $|f_n|\le g$ for all $n$.

Application in Proofs: In the context of the Lebesgue Differentiation Theorem, the dominated convergence theorem is used to justify the interchange of limit and integration when taking the limit of averages of $f$ over shrinking balls.