The definition of dirac delta distribution is that it is a distribution with the property:
$$\int_{-\infty}^{\infty} \delta(x - k)f(x) {\rm d}x = f(k)$$
So if one takes the case $k = 0$ and $f(x) = 1$ then we get the heuristic
$$\delta(x) = \begin{cases} +\infty, & x = 0 \\ 0, & x \ne 0 \end{cases}$$ and satisfies the identity $$\int_{-\infty}^\infty \delta(x) \, dx = 1$$ We have \begin{eqnarray}\int_{-\infty}^{1} \delta(x) \, dx &=& \int_{-\infty}^\infty \delta(x) \, dx - \int_{1}^{\infty} \delta(x) \, dx \\ &=& 1 - 0. \end{eqnarray} I suspect that this is not a correct way to prove this.
For a rigorous proof, you may use the definition of distributions using functional analysis but I suggest this more straight forward argument: let $$f(x)=\begin{cases}1,\ x<1\\0,\ x\ge1\end{cases}$$ then $$\int_{-\infty}^\infty\delta(x)f(x)dx=f(0)=1=\int_{-\infty}^1\delta(x)dx.$$