$\int f d\delta_y = f(y)$ if $f$ is real-valued and $\delta_y$ is the point mass at $y$

Question

Let $\delta_y(A)=1$ if $y\in A$, else $\delta_y(A)=0$. Show if $f:X \rightarrow R$ then $\int f d\delta_y = f(y)$

I proved that $\int f d\delta_y \le f(y)$ but I'm not sure if it's always true that $f(y) \le \int f d\delta_y$ for any $y \in X$.

Answer 1

Step 1 : If $f$ is a simple function, it's obivous.

Step 2 : Suppose $f\geq 0$ and let $\varphi_n$ a sequence of simple function s.t. $\varphi_n\nearrow f$. $$\int f d\delta_y=\lim_{n\to \infty }\int \varphi_nd\delta_y=\lim_{n\to \infty }\varphi_n(y)=f(y).$$

Step 3 : If $f$ is measurable, set $f=f^+-f^-$ with $f^+(x)=\max \{0,f(x)\}$ and $f^-(x)=-\min\{0,f(x)\}$ and apply the previous step.

Answer 2

If $s$ denotes a simple function with $s=\sum_{k=1}^na_k1_{A_k}$ then: $$\int sd\delta_y=\sum_{k=1}^na_n\int1_{A_k}d\delta_y=\sum_{k=1}^na_n\delta_y(A_k)=\sum_{k=1}^na_n1_{A_k}(y)=s(y)$$

If $f$ is nonnegative and measurable and $S_f$ denotes the collection of all nonnegative simple functions $s$ that satisfy $s\leq f$ then: $$\int fd\delta_y=\sup(\{\int sd\delta_y\mid s\in S_f\})=\sup(\{s(y)\mid s\in S_f\})=f(y)$$

Edit: this shows that the integrals of nonnegative functions wrt $\delta_y$ will not take value $+\infty$. Consequently every measurable function $f$ is integrable wrt $\delta_y$.

And applying the definition of integral for such a measurable function $f$ we find:

$$\int fd\delta_y=\int f_+d\delta_y-\int f_-d\delta_y=f_+(y)-f_-(y)=f(y)$$