Is this passage correct?

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How to justify the following passage

$$\int_{j-1}^j \int_{k-1}^{k} \delta (t-\tau) \, dt \, d\tau=\delta_{jk}$$

$j, k\in \mathbb{Z}\; t, \tau \in \mathbb{R} $

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Intuitively, the delta is on the line $t=\tau$. If the cell $[k,k-1]\times[j,j-1]$ doesn't contain that line, the integral is $0$. So now consider the integral

$$\int_{j-1}^j \int_{k-1}^k \delta(t-\tau)dtd\tau = \int_{j-1}^j H(t-\tau)\Bigr|_{k-1}^k d\tau$$ $$ = \int_{k-1}^k H(k-\tau) - H(k-1-\tau) d\tau = \int_{k-j}^{k-j+1} H(\tau')-H(\tau'-1)d\tau'$$

where $H(x)$ is the Heaviside step function. Notice that the integrand is a box of height $1$ between $0\leq \tau' \leq 1$. If $j\neq k$, then the integral integrates over $0$. We only integrate over the nonzero portion of the box when the limits of integration are $0$ and $1$, i.e. $j=k$. Thus the integral evaluates to $\delta_{jk}$