the limit of a convergent sequence of measurable functions is measurable

[[concept]]

Topics

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Real-Valued Functions

Theorem

Let $E\subset \mathbb{R}$ be measurable and $f_n:E\to [-\infty ,+\infty ]$ a sequence of measurable functions. Then the functions

• $g_1(x)={\sup }_n{f}_n(x)$
• $g_2(x)={\inf }_n{f}_n(x)$
• $g_3(x)=\lim {\sup }_{n\to \infty }{f}_n(x)={\inf }_n[{\sup }_{k\ge n}{f}_k(x)]$
• $g_4(x)=\lim {\inf }_{n\to \infty }{f}_n(x)={\sup }_n[{\inf }_{k\ge n}{f}_k(x)]$ Are all measurable functions

Proof

$g_1(x)={\sup }_n{f}_n(x)$ Note that

x \in g_{1}^{-1}(\,(\alpha, \infty]\,) &\iff \sup_{n} f_{n}(x) > \alpha \\ & \iff \exists\, n \text{ s.t. }f_{n}(x) > \alpha \\ &\iff x \in f_{n}^{-1} (\,(\alpha, \infty]\,) \\ &\iff x \in \bigcup_{n} f_{n}^{-1}(\, (\alpha, \infty]\,) \end{align}$$ And each set in this union is [[Concept Wiki/lebesgue measurable\|measurable]] (since each $f_{n}$ is [[Concept Wiki/measurable function\|measurable]]), and thus the preimage of $(\alpha, \infty]$ under $g_{1}$ is also measurable. Thus $g_{1}$ is measurable.

$g_2(x)={\inf }_n{f}_n(x)$ In a similar manner (and using our equivalent intervals), we can check that $x\in g_2^{-1}([\alpha ,\infty ])⟺x\in {\bigcap }_n{f}_n^{-1}([\alpha ,\infty ])$ And since each $f_n^{-1}([\alpha ,\infty ])$ is measurable, we have that the countable intersection is integrable, and therefore that $g_2$ is also measurable.

Finally, $g_3$ and $g_4$ follow immediately from the two proofs above showing measurable closure under both infimums and supremums (since they are the sup and inf of a sequence of measurable functions respectively).

this does not hold for Riemann integration!

The set $\mathbb{Q}\cap [0,1]$ is countable, so enumerate its elements as $r_1,r_2,r_3,\dots$ Then the functions

1 & x \in \{ r_{1},\dots,r_{n} \} \\ 0 & \text{otherwise} \end{cases}$$ Are each Riemann integrable since they are piecewise continuous, but their limit is the indicator function $\mathbb{1}_{\mathbb{Q} \cap [0,1]}$ which is not.

(Riemann integration is not closed under limits)

Extension to Complex-Valued Functions

Theorem

If $f_n:E\to \mathbb{C}$ is measurable for all $n$ and $f_n(x)\to f(x)$ pointwise for all $x\in E$, then $f$ is measurable.

Proof

Note that ${\lim }_{n\to \infty }{f}_n(x)=f(x)⟺{\lim }_{n\to \infty }\mathrm{Re}(f_n(x))=\mathrm{Re}(f(x))\text{ and }{\lim }_{n\to \infty }\mathrm{Im}(f_n(x))=\mathrm{Im}(f(x))$ And since the limit of a convergent sequence of measurable functions is measurable for real-valued functions, we preserve measurability for both $\mathrm{Re}(f(x))$ and $\mathrm{Im}(f(x))$. sums and products of measurable functions are measurable gives us the desired result

References

References

See Also

Mentions

Mentions

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