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.

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