Integration with respect to Dirac measure

Question

The Dirac measure is defined by $$\delta_x(A)= \begin{cases} 1 &\text{if $x \in A$}\\ 0 &\text{if $x \notin A$}\\ \end{cases}$$ Let $f:X\rightarrow \mathbb{R}$ be a function. Can anyone show me why $\int f \, d\delta_x=f(x)$? Any help is appreciated.

Answer 1

Hint If $g=\sum_i c_i 1_{A_i}$ is a step function, then

$$\int g d \delta_x = \sum_i c_i \delta_x(A_i)=\sum_i c_i 1_{A_i}(x)=g(x) $$

Answer 2

Let $g$ denote the constant function being equal to $f(x)$. Then $f=g$ almost surely with respect to $\delta_x$, i.e. the function $f$ is almost surely equal to the constant function taking on the value $f(x)$. This is easily seen since $$ \delta_x(\{g=f\})=\delta_x(\{y\in X\mid g(y)=f(y)\})=1 $$ since $x\in \{g=f\}$ is in that set. Now use that two functions who are identical almost surely have the same integral. That is $$ f=g\quad\delta_x\text{-a.s}\;\Longrightarrow\;\int f\,\mathrm d\delta_x=\int g\,\mathrm d\delta_x=f(x)\int 1\,\mathrm d\delta_x=f(x)\cdot\delta_x(X)=f(x). $$

Answer 3

If this is done for the case where $f$ is everywhere nonnegative, you can probably figure out how to do the rest.

Partition the whole space $X$ into two sets: $\{x\}$ and the complement of $\{x\}$. Look at the simple function $$ g(w)=\begin{cases} f(x) & \text{if }w=x, \\ 0 & \text{if }w\ne x. \end{cases} $$ Then $g\le f$ everywhere and $\int_X g \, d\delta_x=f(x)$. Therefore $\int_X f\,d\delta_x\le f(x)$.

If $h$ is any other simple function $\le f$ then one of the measurable sets on which $h$ is constant contains $x$, and so on that set the value of $h$ cannot exceed $f(x)$. The measure of that set is $1$, and the measure of each of the other measurable sets on which $h$ is constant is $0$. Hence $\int_X h\,d\delta_x\le f(x)$.