If $X$ is absolutely continuous r.v. and $Y=X+1$, can one prove $P(Y=1 \mid X=0)=1$? 
Let's say $X$ is an absolutely continuous random variable, for example $X\sim N(0,1)$, and $Y=X+1$. It is intuitively clear that $P(Y=1 \mid X=0)$ should be $1$, given the strong correlation between the variables.
  However, I can't seem to formally prove this claim. 

Of course, the problem is that we are conditioning on a zero-probability event. I have tried two approaches, but got contradictory results, so I guess there is a mistake hidden somewhere, or that the probability is not even defined.

My attempt(s):
Firstly, let's denote events $A=\{Y=1\}$, $B=\{X=0\}$, $B_n=\{X\in(-\frac1n,\frac1n)\}$ and use the limiting procedure (as written on Wikipedia). Clearly, $B_n \downarrow B$ and we write:
\begin{align}
P(A\mid B)&=\lim_{n\to\infty}\frac{P(A\cap B_n)}{P(B_n)} \\
&= \lim_{n\to\infty}\frac{P(X+1=1, X\in(-\frac1n,\frac1n))}{P(X\in(-\frac1n,\frac1n))}\\
&= \lim_{n\to\infty}\frac{P(X=0, X\in(-\frac1n,\frac1n))}{P(X\in(-\frac1n,\frac1n))}\\
&=\lim_{n\to\infty}\frac{P(X=0)}{P(X\in(-\frac1n,\frac1n))}=0,
\end{align}
as X is continuous, so the numerator is zero for each $n$.
That result intuitively doesn't make any sense, so I tried another way, i.e. to derive the conditional distribution of $Y$ given $X=0$ (using the same limiting procedure).
\begin{align}
P(Y\leq t\mid X=0)&=P(X\leq t-1\mid X=0)\\
&= \lim_{n\to\infty}\frac{P(X\leq t-1, X\in(-\frac1n,\frac1n))}{P(X\in(-\frac1n,\frac1n))}\\
&=
\begin{cases}0,\ t<1 \\\frac12,\ t=1\\ 1,\ t>1\end{cases},
\end{align}
which isn't a distribution function (not right continuous).
However, taking $B_n=\{X\in\left(-\frac1n,0\right]\}$ instead of $\{X\in(-\frac1n,\frac1n)\}$, we get rid of that edge case at $1$ and get a nice distribution function $$P(Y\leq t\mid X=0)=\begin{cases}0,\ t<1 \\ 1,\ t\geq1\end{cases},$$ form which we conclude that $Y\stackrel{a.c.}{=}1$ given $X=0$ and that finally matches our intuition. From there, the beginning claim directly follows.

So, at the end it seems like I managed to give a proof of the claim, but because the results are all over the place, I get the impression that I was merely fishing for proof of my intuition and maybe the wanted probability isn't well defined.
So the question is:

Is the probability $P(Y=1 \mid X=0)$ well defined and if so, is there a formal and unambiguous derivation of its value?


PS Here I did calculations only for $X\sim N(0,1)$ and $Y=X+1$, but I am also interested in the general case of absolutely continuous $X$ and $Y=f(X)$ (maybe a condition such as $f$ being bijective is needed).
 A: I think the correct formalism to resolve this question is the measure-theoretic formulation of conditional expectation, so my answer to your question is to learn about this concept.

For continuous (more precisely, diffuse) random variables, it is meaningless talk about the probability of a single value occurring. More precisely, the probability is always $0$, regardless if the density is quite large or quite small at that point. You simply don't get any information.
The rough idea is that one can introduce a "fudge factor": instead of asking about the probability that $Y=1$ given $X=0$, one could ask something more along the lines of,
$$
\text{what is the conditional probability that $Y$ is close to $1$, given that $X$ is close to $0$?}
$$
For example, one could compute
$$
\mathbb P(1-\epsilon<Y<1+\epsilon\mid -\epsilon<X<\epsilon)
$$
as a stand-in for the (undefined) probability you are trying to compute.

Response to comments below.
Below, you mentioned the continuous form of Bayes' theorem $$\mathbb P(B)=\int_{\mathbb R} \mathbb P(B\mid T=t)\ f_T(t)\ dt.$$
where $B$ is some event. In the context of measure-theoretic probability (which I believe is the correct context to resolve this question, as written above), this expression lacks precise meaning (for more on this, see the edit below) and would instead be written as follows:
$$
\mathbb P(B)=\int_{\mathbb R}\mathbb P(B\mid T)\ d\mu,
$$
where $\mu$ is the probability distribution of $T$ and $B$ is an event which is measurable with respect to $\mathbb P$. Here we are using the Lebesgue integral. We are also assuming that the measure $\mu$ is diffuse, which means that $\mu(\{t\})=0$ for all $t\in\mathbb R$ (this is a formalization of what you mean when you say that $T$ is continuous). In this framework, the density function $f_T(t)$ is given by the Radon-Nikodym derivative of $\mu$ with respect to the Lebesgue measure on $\mathbb R$.
The measure-theoretic definition of $\mathbb P(B\mid T)$ is that it is a random variable belonging to the $\sigma$-algebra generated by $T$, denoted $\sigma(T)$, and satisfying
$$
\mathbb E[\mathbb P(B\mid T); A]=\mathbb P(B\cap A)
$$
for all events $A\in \sigma(T)$. It is a theorem that such a random variable is then uniquely defined up to sets of measure zero (see the beginning of the Martingales chapter of Probability: Theory and Examples by Durrett, for example). In particular, since $\mathbb P(T=t)=0$, the probability of $\mathbb P(B\mid T)$ being equal to $t$ is not determined by the definitions. It could take any value, or even remain undefined if one so desires. In contrast, a well-defined question would be "What is the probability that $\mathbb P(B\mid T)$ lies in a set $A$ of positive measure?" The answer to this question would be
$$
\int_A \mathbb P(B\mid T)\ d\mu.
$$
The key point is that this integral is well-defined, even though $\mathbb P(B\mid T)$ is left unspecified on sets of measure zero. This is because any two functions that disagree on a set of measure zero have the same Lebesgue integral.
TL;DR. Conditional expectation is tricky. Measure theory defines things rigorously, makes conditional probability a random variable, and leaves it unspecified on sets of measure zero.

EDIT: Above I said that the probability $\mathbb P(B\mid T=t)$ lacks precise meaning. Of course, it can be given a precise meaning, as is done rigorously on page 12 of Anderson's classic textbook on multivariate statistical analysis. This is done by defining $\mathbb P(B\mid T=t)$ to be the limit of $\mathbb P(B\mid t\leq T\leq t+\epsilon)$ as $\epsilon\to 0$, if such a limit exists.
