Bounded convergence theorem

Today as I was trying to learn about the Leibniz method of differentiating under the integral sign, I came across the Bounded convergence theorem which happens to be very useful towards proving the one-dimensional case of the Leibniz method. It happens to be simpler to prove than the Lebesgue dominated convergence theorem but its still a very powerful theorem.

First, let’s introduce the notion of convergence in measure which is a useful generalization of the notion of convergence in probability.

Convergence in measure:

f_n \rightarrow f  in measure if \forall \epsilon > 0 ,

\begin{aligned} \mu(\{x \in X: |f_n(x)-f(x)| \geq \epsilon \}) \rightarrow 0 \text{ as } n \rightarrow \infty \end{aligned} 

Bounded convergence theorem:

If \{f_n \} are measurable functions that are uniformly bounded and f_n \rightarrow f in measure as n \rightarrow \infty , then

\begin{aligned} \lim_{n\to\infty} \int f_n dP = \int f dP \end{aligned}

Proof:

Since

\begin{aligned} | \int f_n dP -\int f dP| = | \int (f_n-f) dP| \leq  \int |f_n-f| dP \end{aligned}

we only need to prove that if f_n \rightarrow f in measure and |f_n| \leq M , then \int |f_n-f| dP \rightarrow 0 \text{ as } n \rightarrow \infty . So we have:

\begin{aligned} \int |f_n-f| dP = \int_{|f_n-f|\leq \epsilon} |f_n-f| dP + \int_{|f_n-f| > \epsilon} |f_n-f| dP \end{aligned}

From which we deduce:

\begin{aligned} \int |f_n-f| dP \leq \epsilon + MP \{x: |f_n(x)-f(x)|> \epsilon \} \end{aligned}

However, f_n \rightarrow f in measure so the RHS of the above inequality tends to zero as n \rightarrow \infty . From this the theorem follows.

One thought on “Bounded convergence theorem

  1. Pingback: Differentiating under the integral sign – Kepler Lounge

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