Suppose $\mu$ is a finite measure on the Borel sets of $R$ such that $f(x) = \int_R f(x + t) \mu(dt)$ a.e., show $\mu(\{0\}) = 1$. Problem statement: Suppose $\mu$ is a finite measure on the Borel sets of $R$ such that $f(x) = \int_R f(x + t) \mu(dt)$ a.e., whenever $f$ is real-valued, bounded, and integrable. Show $\mu(\{0\}) = 1$ .
My attempt at a solution: So, to begin with, I don't fully understand the problem statement. Is $\mu(R) < \infty$, or is $\mu(B) <\infty$ for any Borel set of $R$? My idea to show this was to let $f(x) = \chi_{[0,x]}(x)$, which is certainly real-valued, bounded, and integrable. Now, we have
$$\chi_{[0,x]}(x) = \int_R \chi_{[0,x]}(x+t)\mu(dt),$$
and since 
$${\chi _{[0,x]}}(x + t) = \left\{ {\begin{array}{*{20}{c}}
{1,}&{x + t \in [0,x]}\\
{0,}&{else}
\end{array}} \right. = \left\{ {\begin{array}{*{20}{c}}
{1,}&{t \in [ - x,x]}\\
{0,}&{else}
\end{array}} \right. = \chi_{[-x,x]}(t),$$
we have $\chi_{[0,x]}(x) = \int_R \chi_{[-x,x]}(t)\mu(dt) = \mu([-x,x])$. Taking the limit of both sides as $x \to 0$, we get $\chi_{[0,0]}(0) = 1 = \mu(\{0\})$. 
This proof makes me squeamish, because (a) I haven't used the finite measure condition, and (b) this question is in a chapter about $L^p$ spaces, and all I'm really messing around with here is basic integration stuff, and (c) that last line with the limit stuff just makes me uncomfortable. Any help would be much appreciated!
As a side note, I also thought about trying to use the fact that continuous functions with compact support are dense in $L^1$ - is this a more promising idea?
 A: Actually your equalities should be 
$${\chi _{[0,x]}}(x + t) = \left\{ {\begin{array}{*{20}{c}}
{1,}&{x + t \in [0,x]}\\
{0,}&{else}
\end{array}} \right. = \left\{ {\begin{array}{*{20}{c}}
{1,}&{t \in [ - x,0]}\\
{0,}&{else}
\end{array}} \right. = \chi_{[-x,0]}(t)$$, so that ${\chi _{[0,x]}}(x) = \mu([-x,0]) $. Now we have $ \mu([-x,0])=\chi_{[0,x]}(x)=1$ for every x. You do not need to take the limit, the equality works for $x=0$ too.
What you are doing here is considering the function f that has $f(0)=1$ and $f(x)=0$ elsewhere. Then you have that $$1=f(0)=\int_R f(t)\mu(dt)=\mu (\{0\})$$
It is true that the finiteness of $\mu$ almost isn't used, we only needed the much weaker fact that $\mu$ doesn't have an infinite point mass at 0 so that f is integrable. Maybe this is what was meant by "finite measure", although It is not the standard meaning of the term
$$$$
EDIT
Actually I missed the "a.e" in $f(x) = \int_R f(x + t) \mu(dt)$ a.e$\;$. This introduces the possibility that $f(0)\neq \int_R f(t)\mu(dt)=\mu (\{0\})$. To fix this problem, let $g$ be defined by $g(x)=1$. Then 
$$1=g(x)=\int_R g(x + t) \mu(dt)=\mu (R)\;\;a.e$$, so that $\mu (R) =1$. 
Now we also have 
$$0=\chi_{(-\infty,0)}(x)=\int_R \chi_{(-\infty,0)}(x+t) \mu(dt)=\mu ((-\infty,-x))$$ for arbitrarily low $x>0$ (since the equality holds a.e). Thus $\mu(B)=0$ for $B$ of the form $(-\infty, -x)$, and since $(-\infty,0)$ is the union of countably many sets of this form, we have $\mu(-\infty,0)=0$. Similarly, $\mu(0,\infty)=0$. We conclude that we must have $\mu(\{0\})=1$.
Note that In this version of the proof I did use the finiteness of $\mu$ since I assumed that $\chi_{(-\infty,0)}$ is integrable.
