# Dirac measure and counting measure

Let $X$ be a nonempty set. I would like hints in showing the following:
(a) every extended real valued function on $X$ is measurable with respect to the measurable space $(X,2^X)$.
(b) Let $x\in X$. Let $\delta_x$ be the Dirac measure at $x$ on $2^X$. Then two functions on $X$ are equal almost everywhere $[\delta_x]$ if and only if they take the same value at $x$.
(c) Let $\alpha$ be the counting measure on $2^X$. Then two functions on $X$ are equal almost everywhere $[\alpha]$ if and only if they take the same value at every point in $X$.

Following Davide's hints,

(a) Let $\mathcal{B}=2^X$. Then $2^X$ is a sigma algebra and by definition the if $(X,2^X)$ is a measure space, then $f:X\to \overline{\mathbb{R}}$ is measurable since the the set $\{x\in X : f(x) < c\} \in 2^X$ for every $c\in \mathbb{R}$.

(b) Here, it suffices to show that the sets which have measure zero are the sets which does not contain $x$. So suppose $f=g$ a.e. then there is a set $N$ with measure zero such that $A=\{x\in X : f(x)\ne g(x)\} \subset N$. But these are precisely those points which don't contain $x$. So $f(x)=g(x)$ on $N^c$.

(c) By definition of the counting measure , the empty set is the only set with counting measure zero. So $f=g$ a.e. $\Leftrightarrow f(x)=g(x)$ at every point in $X$.

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Yes it work (if you already know that to check that a function with values in $\overline{\mathbb R}$ is measurable we only need to show the measurability of the sets $\{f<c\}$, $c\in\mathbb R$. Otherwise, you have to take an arbitrary set of the Borel $\sigma$-algebra on $\overline{\mathbb R}$. – Davide Giraudo Feb 12 '12 at 19:43
@DavideGiraudo: How about (b) and (c)? – KKK Feb 12 '12 at 20:19
All of these follow immediately from definitions... – Carl Mummert Apr 13 '12 at 19:17

(a) If $f\colon X\to\overline{\mathbb R}$ (with any $\sigma$-algebra $\mathcal B$) then for each $B\in\mathcal B$, $f^{-1}(B)$ is a subset of $X$.
(b) The sets which have measure $0$ are all the sets which doesn't contain $x$.
(c) Show that the only set which have counting measure $0$ is the emptyset.
Both directions, since to know what "almost everywhere" mean in these particular cases, you have to know what are the sets of measure $0$. – Davide Giraudo Feb 12 '12 at 19:19