is it a single variable probability density function? $$f(x)=\left\{ \begin{array}{ll} e^{3x} , &\rm{~~if~~} x \leq 0\\ 1 - \frac{2}{3}x ,&\rm{~~if~~} 0 < x \leq 1\\ 0, &\rm{~~if~~} x > 1\\ \end{array}\right.$$
I checked, the definite integral of f(x) from -infinity to infinity is 0. $f(x)\geq 0$
it's limits to infinity and -infinity equal to 0. It is continuous from left.
There is a 'bump' in the function, because $f(1)=1/3$ and $f(x>1)=0$ So is it not a problem, because it only has to be continuous from the left, is it correct?
Thank you
A density function is a non-negative measurable function whose integral is $1$. Continuity is not required. (You are probably confusing density function with distribution function). Here the integral is $\frac 1 3 +(1-\frac 1 3) +0=1$ so $f$ is a density function.