Minimum of a constant and a random variable Let $X$ be a random variable and I now assume for simplicity that it is uniformly distributed on $[0,1]$.
Now fix $t \in [0,1]$ and let $Y_t$ be the random variable $Y_t:=\min \{t,X\}$.
I have problems computing for example the density
Computing the cdf is easy: 
Of course $P(Y_t \leq y)= 1- P(t>y,X>y)$. Now I tried to split this apart as if $t$ was just random variable independet of $X$ and this gives $P(Y_t \leq y)=y$ if $y<t$ and $1$ if $y\geq t$. 
However, I feel that taking the derivative is not the appropriate her since $Y_t$ attains
the value $t$ with positive probability. So $Y_t$ seems to be a random variable that is neither a discrete nor a continuous random variable. So maybe a density does not make sense.
However, in order to be able to compute expectation and so on, I would need a density.
In fact I guess everything becomes easy if I knew on which space $Y_t$ actually lived.
However, I have problems formalizing that. Can anybody help.
Maybe for the uniform distribution everything is easy by direct arguments. But I would like to see a general approach which is valid for $X$ having an continuous distribution.
 A: Your approach is correct. The distribution of $Y_t$ is partly absolutely continuous with respect to the Lebesgue measure, and partly discrete. A formal notation for the distribution $P_{Y_t}$ of $Y_t$ is
$$
\text{d}P_{Y_t}(x)=\mathbf{1}_{[0,t]}(x)\text{d}x+(1-t)\delta_t(\text{d}x).
$$
But this is only a notation. More to the point, for every Borel subset $B$,
$$
P(Y_t\in B)=\text{Leb}(B\cap[0,t])+(1-t)\mathbf{1}_B(t).
$$
And to compute expectations, for every bounded measurable function $u$, use
$$
E(u(Y_t))=\int_0^tu(x)\text{d}x+(1-t)u(t).
$$
In the general case, if $X$ has density $f_X$, one gets
$\text{d}P_{Y_t}(x)=f_X(x)\mathbf{1}_{x<t}\text{d}x+P(X\ge t)\delta_t(\text{d}x),$ hence
$$
P(Y_t\in B)=P(X\in B\cap(-\infty,t))+P(X\ge t)\mathbf{1}_B(t),
$$
and
$$
E(u(Y_t))=\int_{-\infty}^tu(x)f_X(x)\text{d}x+u(t)\int_t^{+\infty}f_X(x)\text{d}x.
$$
A: Didier's and Yuval's answers are the correct formal way to compute the expectation.
More informally: A (purely) continuous random variable that takes values in some interval (say [0,1]) has zero probability of taking a particular value. Which is the same as saying that the distribution function is continuouos (typically piece-wise derivable), which is the same as saying that the density function is a "true" function (i.e. not including Dirac deltas).
By contrast, a discrete variable has finite non-zero probabilities of taking a (discrete) set of values, and its distribution function is discontinuous, piecewise constant.
A uniform variable in [0,1] is clearly continuous. But $Y_t$ has a finite probability of taking the value $t$, hence the density will exhibit a Dirac delta there, and the distribution function will have a discontinuity.
Hence, as Yuval points out, $Y_t$ can be seen as a mixture (a convex combination) of two random variables: a continuous one (uniform in [0,t]) and a discrete variable (a constant, actually, with value $t$). The mix of two variables gives a density function that is the same linear combination of the two densities (here a truncated uniform, and a dirac delta) (you should graph this), and the same holds (by linearity) for the expectations.
A: Some probability distributions are discrete, some are continuous, and some are neither. If you want to use one integral-like formula for all, you can use the Riemann-Stieltjes integral.
Alternatively, every "reasonable" random variable (see discussion in the comments below) is a convex combination of a discrete random variable (formed by taking all its atoms, points with positive probability) and a continuous random variable (the rest). Whatever it is that you're calculating is a convex combination of the same thing on both parts, and for the latter you already know how to calculate things like expectation.
For example, suppose that $0 \leq t \leq 1$. Then $Y_t$ is a convex combination of the constant $t$ r.v. (with probability $1-t$), and of a uniform $[0,t]$ r.v. (with probability $t$). Hence its expectation is $$(1-t)t + t(t/2) = t-t^2/2.$$
