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I just started studying Rudin's RCA,and I find that the codomains of measurable functions appearing in statements vary.Sometimes $[0,∞]$,sometimes$[-∞,∞]$.

Since a measurable function is a function from a measurable space to a topology space in Rudin's book,different codomains have different topology.

I have difficulty to understand the topology of $[0,∞]$,how do we define it usually?I basically know nothing about topology apart from definition of topology space.

Is a measurable function with codomain $[0,∞]$ also measurable when regarded as a function with codomain $[-∞,∞]$?And does the converse hold?(consider the nonnegative measurable function with codomain $[-∞,∞]$)

Thanks for your help.

Mugenen
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In $(0,\infty]$ we use the subspace topology. As $(0,\infty]$ is an open subset of $[-\infty,\infty]$, the open sets in the subspace topology of $(0,\infty]$ are open sets of $[-\infty,\infty]$ and all the open sets of $[-\infty,\infty]$ contained in $(0,\infty]$ are open sets in the subspace topology of $(0,\infty]$.

Martín-Blas Pérez Pinilla
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$[-\infty,\infty]$ is equipped with the order topology $\tau$ and (Borel) $\sigma$-algebra $\mathcal B=\sigma(\tau)$.

If $A\subseteq[-\infty,\infty]$ and $i:A\to[-\infty,\infty]$ denotes its inclusion then $A$ is equipped with subspace topology $i^{-1}(\tau)=\{A\cap U\mid U\in\tau\}$.

This subspace topology induces at first hand the $\sigma$-algebra $\mathcal B_A:=\sigma(i^{-1}(\tau))$ but fortunately it can be shown that:$$\mathcal B_A=\sigma(i^{-1}(\tau))=i^{-1}(\sigma(\tau))=i^{-1}(\mathcal B)=\{A\cap B\mid B\in\mathcal B\}$$The second equality is essential here.

Now let $(\Omega,\mathcal A)$ be a measurable space and let $f:\Omega\to[-\infty,\infty]$ be a measurable function - that is $f^{-1}(\mathcal B)\subseteq\mathcal A$ - with $f(\Omega)\subseteq A$.

Then the question rises: is $f$ also measurable if it is regarded as a function $\Omega\to A$?

Fortunately the answer is: "yes".

Observe that $f(\Omega)\subseteq A$ allows us to write $f=ig$ where $g:\Omega\to A$ is prescribed by $\omega\mapsto f(\omega)$.

Then the actual question is now: is function $g:\Omega\to A$ measurable?

Now observe that: $$g^{-1}(\mathcal B_A)=g^{-1}(i^{-1}(\sigma(\tau)))=(ig)^{-1}(\sigma(\tau))=f^{-1}(\sigma(\tau))=f^{-1}(\mathcal B)\subseteq\mathcal A$$which is exactly what we need for measurability of $g$.

If conversely we start with a measurable function $g:\Omega\to A$ then we might wonder whether it is measurable if it is looked at as a function $\Omega\to[-\infty,\infty]$.

That question can be rephrased as: is function $f:=ig:\Omega\to[-\infty,\infty]$ measurable?

Again the answer is "yes".

For this note that $i:A\to[-\infty,\infty]$ is measurable simply because $i^{-1}(\mathcal B)=\mathcal B_A$.

Then $f=ig$ can be recognized as composition of two measurable functions and a composition of measurable functions is measurable.

drhab
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  • Thanks.But the definition of measurable function in Rudin's book is a map from measurable space to topological space such that preimage of an open set is a measurable set.So maybe you prove a stronger statement since you use open sets to generate Borel sets.Is my understanding right?I'm uncertain about it.. – Mugenen Mar 08 '18 at 23:22
  • Commonly a function is measurable if preimage of a measurable set is measurable. If it is defined as in Rudin then it is only demanded that $f^{-1}(\tau)\subseteq\mathcal A$ where $\tau$ denotes a topology on the codomain and $\mathcal A$ a $\sigma$-algebra. Then $\sigma(f^{-1}(\tau))\subseteq\mathcal A$ and it can be shown that $\sigma(f^{-1}(\tau))=f^{-1}(\sigma(\tau))$. So actually there is no essential difference in demanding $f^{-1}(\tau)\subseteq\mathcal A$ or demanding $f^{-1}(\mathcal B)\subseteq\mathcal A$ where $\mathcal B$ denotes the $\sigma$-algebra generated by $\tau$. – drhab Mar 09 '18 at 08:37