Can any measure be made into a bounded measure? Is it possible to derive a bounded measure from any measure on a measure space?
For example can the Lebesgue measure be made into a probability measure?
 A: Let $\mu$ be a measure and $f\in L^{1}(d\mu)$, $f\ge0$, and $\int fd\mu\neq0$ then $fd\mu$ is a bounded measure which can be normalized to be a probability measure. Such an $f$ can be found if the measure is semifinite. For Lebesgue measure we can take the measure $ce^{-\lvert x\rvert^{2}}dm$ where $c$ is a normalizing constant. The idea here was simply to multiply by a weight to make the measure a probability measure. To see that it can't be done for all measures consider the trivial measure.
A: The question asking "Is it possible to derive a bounded measure from any measure on a measure space" I understand as follows: For a fixed measure $\lambda$, does there exist a probability measure $\mu$ which is equivalent to $\lambda$(under equivalence we understand that $\lambda(X)=0$ iff $\mu(X)=0$).  
Then the answer for a $\sigma$-finite non-trivial measure is yes. For non-$\sigma$-finite measures the answer(in general) is no. Indeed, let consider the measure $\lambda$  constructed in [Baker R.,  ``Lebesgue measure" on $\mathbb{R}^{ \infty}$.
II. \textit{Proc. Amer. Math. Soc.} vol. 132, no. 9, 2003, pp. 2577--2591]. It is well known that the restriction of the $\lambda$ to the  $\sigma$-algebra of Borel subsets of $R^{\infty}$(for this restriction we preserve the same notation $\lambda$) is a non-$\sigma$- finite Borel measure which is not concentrated on a union of countable family of compact sets in $R^{\infty}$. If we   assume that there is a Borel probability measure $\mu$ which is equivalent to the measure $\lambda$ then we get that the measure $\lambda$ like $\mu$ is concentrated on  a union of countable family of compact sets in $R^{\infty}$. This is the contradiction. 
