# Limit of measurable functions is measurable

During my revision for my measure theory exam, I had to demonstrate the following proposition: If $f_n$ is a sequence of measurable functions with respect to a sigma-algebra, $\mathcal{A}$, where $f_n: \Omega \rightarrow \mathbb{R}$, and $f_n$ converges pointwise to $f$ on $\Omega$, then $f$ is measurable with respect to $\mathcal{A}$.

After some reflection, I came up with a nice proof based on the fact that $\mathcal{A}$ is closed under countable unions and countable intersections.

Proof:

$\forall c \in \mathbb{R}$, we have:

$latex \bigcap_{m=1}^\infty \bigcup_{n=m}^\infty f_n^{-1}((c,\infty))= \limsup_{n\to\infty} f_n^{-1}((c,\infty)) \in\mathcal{A} \quad (1) &s=1$

$latex \bigcup_{m=1}^\infty \bigcap_{n=m}^\infty f_n^{-1}((c,\infty))=\liminf_{n\to\infty} f_n^{-1}((c,\infty)) \in \mathcal{A} \quad (2) &s=1$

From this it follows that both:

$latex \limsup_{n\to\infty} f_n \quad (3) &s=1$

$latex \liminf_{n\to\infty} f_n \quad (4) &s=1$

are measurable.

Further, given that $\lim_{n\to\infty} f_n = f$:

$f=\limsup_{n\to\infty} f_n = \liminf_{n\to\infty} f \quad (5)$

As a result, we may conclude that $f$ is measurable.

Although this proof may appear simple, this result is very important because all Riemann integrable functions are Lebesgue measurable, and the Lebesgue measurable sets form a sigma-algebra.Therefore the point-wise limit of Lebesgue measurable functions, if it exists, is Lebesgue measurable.

However, the point-wise limit of Riemann integrable functions, if it exists, is not necessarily Riemann integrable. The math stackexchange has some examples.