# definition of $f$ being measurable

I am having troubles understanding the definition of a function being measurable. I have that for a measure space $(\Omega, \mathcal{F}, \mu)$ a function $f: \Omega \to \mathbb{R}$ is measurable if $\{x | f(x) > c\} \in \mathcal{F}$ for every $c \in \mathbb{R}$

Now, I looked at an example which showed that if $f$ is an increasing function, then $f$ is measurable. It stated that since $f$ is increasing, the set $\{ x | f(x) > c\}$ is an interval, and since intervals are Lebesgue measurable we have that $f$ is Lebesgue measurable. I understand that $\{ x | f(x) > c\}$ is an interval, and intervals are Lebesgue measurable, but I don't see why this implies that $f$ is Lebesgue measurable. From the definition we require that $\{x | f(x) > c\} \in \mathcal{F}$ - What does this mean exactly? I thought it just meant that it's an element of the $\sigma$ algebra $\mathcal{F}$.

• Elements of the Sigma algebra are called measurable sets. – Lionel Ricci Apr 12 '16 at 12:16
• When your set is part of the sigma algebra F, it , by definition, means that it is measurable. Elements of F , and only elements of F are measurable sets. – астон вілла олоф мэллбэрг Apr 12 '16 at 12:18
• I defined a measurable set $A$ as one such that $\mu(T) = \mu(T\cap A) + \mu(T \backslash A)$ for any $T \subset \Omega$ we then say $A$ is $\mu$ measurable, I don't see how this coincides with $A \in \mathcal{F}$ – lebesgue Apr 12 '16 at 12:18
• It is enough to show that preimagaes of open sets are measurable, but open sets of reals are countable union of disjoint open intervals and you can generate every open interval using a particular type say the type you have in your problem. So sometimes we just reduce what needs to be checked to just the preimage of a certain type of open set is measurable. – akech Apr 12 '16 at 12:24

In the example you referred to, $\Omega$ is $\mathbb R$, $\mathcal F$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb R$, and $\mu$ is Lebesgue measure on $\mathbb R$. So, the statement that $\{ x \mid f(x) > c \} \in \mathcal F$ means (in this example) that $\{ x \mid f(x) > c \}$ is a Lebesgue measurable subset of $\mathbb R$. If $f$ is increasing, then $\{ x \mid f(x) > c \}$ is an interval, and any interval is Lebesgue measurable.
• I just don't see how $\mathcal{F}$ is the $\sigma-$algebra of Lebesgue measurable subsets of $\mathbb{R}$. We just denoted the measure space $(\Omega,\mathcal{F},\mu)$ where $\Omega$ is a set, $\mathcal{F}$ is a sigma algebra of subsets of the set $\Omega$ and $\mu$ is a measure on $\mathcal{F}$ – lebesgue Apr 12 '16 at 12:24
• I'm guessing that in your example they simply assume that $\mathcal F$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb R$. The statement that $f$ is "increasing" doesn't make sense in a general measure space. – littleO Apr 12 '16 at 12:32