Test post

2026-07-25T00:00:00+00:00

Monotone Convergence Theorem. If $f_n$ are nonnegative measurable functions and $f_n \uparrow f$ pointwise, then

$$\int_X f \, d\mu = \lim_{n \to \infty} \int_X f_n \, d\mu$$

Proof</a></h2>
Let $L = \lim_{n \to \infty} \int_X f_n \, d\mu$, which exists because the integrals form an increasing sequence. Since $f_n \le f$ pointwise,
$$L \le \int_X f \, d\mu.$$
For the reverse inequality, fix a nonnegative simple function $s \le f$ and $0 < c < 1$. Define
$$E_n = \{x \in X : f_n(x) \ge c s(x)\}.$$
The sets $E_n$ increase and cover $X$: if $s(x) > 0$, then $f_n(x) \uparrow f(x) \ge s(x) > c s(x)$, so $x$ eventually lies in $E_n$. On $E_n$, $c s \le f_n$, hence
$$c \int_{E_n} s \, d\mu \le \int_X f_n \, d\mu \le L.$$
Because $s$ is simple and $E_n \uparrow X$, continuity from below of measures gives
$$\lim_{n \to \infty} \int_{E_n} s \, d\mu = \int_X s \, d\mu.$$
Thus $c \int_X s \, d\mu \le L$. Letting $c \uparrow 1$ gives $\int_X s \, d\mu \le L$. Finally, the Lebesgue integral is the supremum of the integrals of nonnegative simple functions below $f$, so
$$\int_X f \, d\mu \le L.$$
Together with the first inequality, this proves the result.
fn main() { let greeting = "hello, world"; println!("{greeting}"); }</code></pre>

Arthur Queiroz Moura

Test post