Is $\int^{\infty}_{-\infty}\delta(x-x_0)f(x) \, dx = f(x_0)$ sufficient to define delta distributions?

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Most of the sources start introductory section of the delta distributions by defining \begin{eqnarray} \delta(x-x_0)&=&\begin{cases} \infty, & \text{if $x=x_0$}.\\ 0, & \text{otherwise}. \end{cases} \\ \int^{\infty}_{-\infty}\delta(x-x_0) \, dx &=& 1 \end{eqnarray}

then the statement follows that, \begin{eqnarray} \int^{\infty}_{-\infty}\delta(x-x_0)f(x) \, dx &=&\lim_{\epsilon \to 0} \left(\int^{x_0-\epsilon}_{-\infty}\delta(x-x_0)f(x) \, dx + \int^{x_0+ \epsilon}_{x_0-\epsilon}\delta(x-x_0)f(x) \, dx \\+ \int^{\infty}_{x_0+\epsilon}\delta(x-x_0)f(x) \, dx \right) \\ &=& f(x_0)\int^{x_0+ \epsilon}_{x_0-\epsilon}\delta(x-x_0)f(x) \, dx \\ &=& f(x) \end{eqnarray}

Why not define last equation to define delta distributions and use first and second statements as corollaries?