Conditional Entropy: if $H[y|x]=0$, then there exists a function $g$ such that $y=g(x)$ Suppose that the conditional entropy $H[Y|X]$ between two discrete random variables $x$ and $y$ is zero.
Show that, for all values of $x$ such that $p(x) > 0$, the variable $y$ must be a function of $x$, in other words, for each $x$ there is only one value of $y$ such that $p(y|x)\ne 0$
Therefore, if $H[y|x]=0$, then there exists a function $g$ such that $y=g(x)$.

Would it be sufficient to prove this by looking at Venn Diagram and noticing that if $H[Y|X]=0$, then:
$$H[y] = H[x] - H[x|y]$$
Therefore, $H[y|x]=0$ if and only if the value of $y$ is completely determined by the value of $x$.
I want to say that $H[x|y]$ depends only on $x$ variable and not on variable $y$... but not sure how to prove it.
I guess, the question is: how to make such a bold statement like above. iff $H[y|x]=0$,  $y$ is a function of $x$!
 A: Write $\mathcal{X}$ and $\mathcal{Y}$ for the range $X$ and $Y$, respectively. Write $H[Y|X]$ as
$$ H[Y|X] = \sum_{x\in\mathcal{X}} \sum_{y\in\mathcal{Y}}  \left( -p(x) p(y|x) \log p(y | x) \right) \tag{1}$$
Here, $0 \log 0$ is regarded as $0$ as usual. Now assume that $H[Y|X] = 0$. Since each summand of $\text{(1)}$ is non-negative, this implies that each summand vanishes, and so, we get
$$ p(y|x) \log p(y | x) = 0 \quad \text{whenever} \quad p(x) > 0. $$
But since the function $t \mapsto t \log t$ on $[0, 1]$ has exactly two zeros $t = 0$ and $t =1$, this implies that
$$ p(y|x) \in \{0, 1\} \quad \text{whenever} \quad p(x) > 0. $$
Still assuming that $x$ satisfies $p(x) > 0$, the identity $\sum_{y\in\mathcal{Y}} p(y|x) = 1$ ensures that there exists a unique $y \in \mathcal{Y}$ such that $p(y|x) = 1$ holds. In such case, we write $y = g(x)$. We also remark that $p(y'|x) = 0$ if $y' \neq g(x)$.
To extend $g$ to a function on all of $\mathcal{X}$, assign to $g(x)$ an arbitrary value in $\mathcal{Y}$ whenever $p(x) = 0$. (Such assignment will never harm the argument.) This yields a function $g : \mathcal{X} \to \mathcal{Y}$. We claim that

Ciaim. $Y = g(X)$ with probability one.

Indeed, recall that $g$ is defined so as to satisfy
$$ p(y|x)
= \mathbf{1}_{\{y = g(x)\}}
:= \begin{cases}
1, & \text{if $y = g(x)$} \\
0, & \text{if $y \neq g(x)$}
\end{cases} \tag{2} $$
whenever $p(x) > 0$. (Here, indicator function notation is used.) From this,
\begin{align*}
\mathbb{P}(Y = g(X))
&= \mathbb{E}[\mathbf{1}_{\{Y = g(X)\}}] \\
&= \sum_{x\in\mathcal{X}} \sum_{y\in\mathcal{Y}} p(x,y) \mathbf{1}_{\{y = g(x)\}} \\
&= \sum_{x\in\mathcal{X}} p(x)\sum_{y\in\mathcal{Y}} p(y|x) \mathbf{1}_{\{y = g(x)\}} \\
&\stackrel{\text{(2)}}= \sum_{x\in\mathcal{X}} p(x)\sum_{y\in\mathcal{Y}} p(y|x) \\
&= \sum_{x\in\mathcal{X}} \sum_{y\in\mathcal{Y}} p(x, y) \\
&= 1.
\end{align*}
A: Based on the definition of Entropy:
$$H[y|x] = - \sum_{x_i} \sum_{y_j} p(x_i, y_j) \ln p(y_j | x_i)$$
Considering the property of probability, we can obtain that:
$$0 \le p(y_j|x_i) \le 1$$
$$0 \le p(x_i, y_j) \le 1$$
Therefore, we can see that:
$$-p(x_i, x_j) \ln p(y_j | x_i) \ge 0$$
when:
$$0 < p(y_j|x_i) \le 1$$


when:
$$p(y_j|x_i) = 0$$
provided with the fact that:
$$\lim_{p \to 0} p \ln p = 0$$
we can see that:
$$-p(x_i,y_j) \ln p(y_j | x_i) = -p(x_i) p(y_j|x_i) \ln p(y_j|x_i) = 0$$
(here we view p(x) as a constant).
Hence for an arbitrary term in the equation above, we have proven that it cannot be less than 0.  In other words, if and only if every term of $H[y_j|x_i]$ equals 0, $H[\mathbf{y}|\mathbf{x}]$ will equal 0.
Therefore, for each possible value of random varuable x, denoted as $x_i$:
$$-\sum_{y_i} p(x_i, y_i) \ln p(y_i|x_i) = 0$$
If there are more than one possible value of random variable "y given $x_i$",  $p(y_i|x_i)$, such that: 
$$p(y_i|x_i) \ne 0$$
(Because $x_i$, $y_i$ are both "possible", $p(x_i, y_i)$ will also not equal to 0), constrained by:
$$0 \le p(y_i|x_i) \le 1$$
and 
$$\sum_j p(y_i|x_i)=1$$
there should be at least two values of y that satisfy:
$$0 < p(y_j|x_i) < 1$$
which ultimately leads to:
$$-\sum_{y_i} p(x_i, y_i) \ln p(y_i|x_i) > 0$$
Therefore, for each possible value of x, there will only be one y such that:
$$p(y|x) \ne 0$$
In other words, y is determined by x.
Note: If y is a function of x, we can obtain the value of y as soon as observing a x.  Therefore we will obtain no additional information when observing a $y_j$ given an already observed x.
