I assume that $\varphi:[0,1]\to\mathbb R$.
Definition of convex is: For each $\alpha\in[0,1]$ and for each $x$, $y$ the inequality $\varphi(\alpha x+(1-\alpha)y) \le \alpha \varphi(x) + (1-\alpha) \varphi(y).$
Try to use the following measure: $\mu(A) = \alpha \chi_{\{x\}}(A) + (1-\alpha) \chi_{\{y\}}(A)$ and the function $f(x)=x$.
Here $\chi_B$ denotes the characteristic function of the set $B$ (a.k.a. indicator function). The measure $\mu$ is basically a convex combination of two Dirac measures $\delta_x$ and $\delta_y$, i.e. $\mu(A)=\alpha \delta_x(A)+ (1-\alpha) \delta_y (A).$
To prove that this implies convexity it suffices to show that for any function $g$ we have $\int g \mathrm{d}\mu = \alpha g(x) + (1-\alpha) g(y).$
EDIT: I understood the question as follows: Suppose that Jensen's inequality holds for arbitrary measure and arbitray $\varphi$, $f$. Then prove convexity. Srivatsan's answer is better, since he only uses the usual (Lebesgue) measure - which was probably the original intention of the question. (Although I have a simpler function $f$...)