Integral leading to step function of zero argument Consider the following integral with real $a>0$
$$ I(a) = \int_0^{\infty} \mathrm{d}x~\delta(a-x) \Theta(x-a) f(x) $$
with $f(x)$ a function such that $f(a) \ne 0$, $\delta$ the Dirac function and $\Theta$ the unit step function. 
A naive solution appears to be
$$I(a) = \Theta(0) f(a) $$.
How to make sense of this result? What to take for $\Theta(0)$?
How to evaluate integral $I(a)$ in a more rigourous way?
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{J}:\mathbb{R}^{2}\setminus
\braces{\pars{x,y}\ \mid\ x = 0\ \vee\ x = y} \to R}$.

\begin{align}
\mrm{J}\pars{a,b} & \equiv
\int_{0}^{\infty}\delta\pars{a - x}\Theta\pars{x - b}\,\mrm{f}\pars{x}\,\dd x =
\bracks{\int_{0}^{\infty}\delta\pars{a - x}\,\dd x}
\Theta\pars{a - b}\,\mrm{f}\pars{a}
\\[5mm] & =
\Theta\pars{a}\Theta\pars{a - b}\,\mrm{f}\pars{a}\implies
\left\{\begin{array}{rcl}
\ds{\mrm{J}\pars{a,a^{-}}} & \ds{=} & \ds{\Theta\pars{a}\,\mrm{f}\pars{a}}
\\[1mm]
\ds{\mrm{J}\pars{a,a^{+}}} & \ds{=} & \ds{0}
\end{array}\right.
\end{align}
