First, we can speak of a causal filter or a causal function (or sequence, or signal).
In general, a filter is causal if its output at present time ($n$) never depends on the input at future times ($n+m$, with $m>0$). Let us restrict to LTI filters (and we assume discrete time) - so that the filter is fully specified by an impulse response function $h(n)$.
In that case, the above property can be concisely stated as follows:
- A LTI discrete-time filter is causal iff $h(n)=0$ for $n<0$
This motivates the definition of a causal signal.
- A discrete-time function (signal, sequence) $g(n)$ is causal iff $g(n)=0$ for $n<0$
Notice that this later definition does not involve filters (it's just motivated by them). And notice that the two can be combined in:
- A LTI discrete-time filter is causal iff its response function $h(n)$ is a causal function.
In your assertion regarding the second link ("it is causal because both $x_n$ and $y_n$ are $0$ for $n<0$") you seem to be confusing both meanings. To determine that a filter is causal, one does not look for the causality of inputs or ouputs ($x_n$ and $y_n$) but for the causality of the response function $h_n$
Further, instead of $h(n)$ we can work with its Z-transform $H(z)$; we have $y[n] = x[n] \star h[n] \implies Y(z)=H(z) \, X(z)$. But, remember the relation "signal" $\leftrightarrow$ "Z transform" is not one-to-one unless the ROC (region of convergence) is also specified. A single $H(z)$ can have several corresponding $h[n]$("anti-transform"), for different ROCs. Alternatively, instead of giving a ROC, we might be given a causal (or anticausal) condition. In particular, if we are given a (rational) $H(z)$ and we are told that $h(n)$ is causal, then the ROC must extend outwards from the biggest pole. For an explanation of this, see any Signal Processing textbook, or here.
In your example, $Y(z)= z X(z)$, so $H(z)=z$. To analyze this you can reason in two ways:
1) In terms of zeros and poles. $H(z)$ has a zero at $z=0$ and a pole at infinity. Hence, because there is a single ROC (all the plane), and it extends inwards from the pole, hence there can be only one valid $h(n)$ which must be anti-causal.
(I insist: here you could deduce from $H(z)$ that the filter was anti-causal; but normally you can't; say, if $H(z)=z/(z-1)$ you'd have two possible $h[n]$, one causal, one anticausal).
2) Explicitly, formally. By inspection, $H(z)=\sum_{n} h(n) z^{-n} = z \implies h(n)=\delta(n+1)$ Which is anticausal.
Best Answer
Let $ (Lx)(t) = \int_{-\infty}^t x(\tau)d \tau$.
You want to show that if $x_1,x_2$ are such that $x_1(s) = x_2(s)$ for $s \le t$, then $(Lx_1)(t) = (L x_2)(t)$.
You have $(Lx_1)(t) = \int_{-\infty}^t x_1(\tau)d \tau = \int_{-\infty}^t x_2(\tau)d \tau = (L x_2)(t)$.
Hence the system $L$ is causal.
Alternatively, albeit the use of distributions is unnecessary, we can look at the response of the system when subjected to an impulse input. You need to verify that the system is linear and time invariant first, then note that if $t < 0$, we have $h(t) = (L \delta)(t) = 0$.