This is one of those things that is helpfully studied using an example. A very nice example for this issue is the "wandering block". Informally, the wandering block is the sequence of indicator functions of $[0,1],[0,1/2],[1/2,1],[0,1/4],[1/4,1/2],[1/2,3/4],[3/4,1]$, etc. More explicitly, it is the sequence $g_n(x)$ which comes about from enumerating the "triangular array" $f_{j,k}(x)=\chi_{[j2^{-k},(j+1)2^{-k}]}(x)$, where $k=0,1,\dots$ and $j=0,1,\dots,2^{k}-1$.
The sequence $g_n$ converges in measure to the zero function. You can see this as follows. Given $n$, write $g_n=f_{j,k}$, then $m(\{ x : |g_n(x)| \geq \varepsilon \})=2^{-k}$ for any given $\varepsilon \in (0,1)$. Since $k \to \infty$ as $n \to \infty$, this measure goes to $0$ as $n \to \infty$.
On the other hand, the sequence $g_n(x)$ does not converge at any individual point, because any given point is in infinitely many of these intervals and also not in infinitely many of these intervals. Thus the sequence $g_n(x)$ contains infinitely many $1$s and infinitely many $0$s, and so it cannot converge.
On an infinite measure space, there is an example for the other direction: $f_n(x)=\chi_{[n,n+1]}(x)$ on the line converges pointwise to $0$ but does not converge in measure, since $m(\{ x : |f_n(x)| \geq \varepsilon \})=1$ for $\varepsilon \in (0,1)$. A corollary of Egorov's theorem says that this is impossible on a finite measure space.
On a related note, the wandering block example also shows a nice, explicit example of the theorem "if $f_n$ converges in measure then a subsequence converges almost everywhere". Here, for any fixed $j$, the sequence $h_k=f_{j,k}$ (defined for sufficiently large $k$ that this makes sense) converges almost everywhere.
I believe the following answers your question, even though it might not be immediately apparent to you.
Let $(\Omega,\mathcal{B},\nu)$ and $(M,\mathcal{A},\mu)$ be measure spaces (let's say $\Omega$ is actually a probability space) and let $\alpha:\Omega\rightarrow M$ be a measurable function. We can define a new probability measure on $M$, called the law or distribution of $\alpha$, as follows: the measure of any set $A\in\mathcal{A}$ is now defined to be the probability (according to $\nu$) that $\alpha(\omega)\in A$, in other words $$ \mathbb{P} (A) := \nu(\{\omega \in \Omega \vert \alpha(\omega) \in A\}).$$ This is well defined precisely because we assumed that $\alpha$ is measurable. Another way of writing this is to say that $\mathbb{P}(A) := \nu( \alpha^{-1} (A))$; yet another way of writing the same thing is $\mathbb{P} := \nu \circ \alpha^{-1}$. Another name for this measure is the pushforward of $\nu$ onto M with respect to $\alpha$, or just the pushforward to be short. (I'm saying all of this because the probability literature uses all of this interchangeably so you may as well know that ahead of time.)
Now let me attempt to convince you that the law of $\alpha$ is the thing that you actually want, rather than something to do with Dirac deltas.
Firstly, you say that your motivation is that you want to "place uncountably many points" in $M$ according to $\alpha$. Well, suppose that due to measuring precision, we cannot actually check "how many" of these points land at any specific $\theta\in M$ - rather, we have to look at a region of the space $A$ which has finite measure, and see what "fraction" of the points have ended up in $A$. But this is precisely $\nu(\{\omega \in \Omega \vert \alpha(\omega) \in A\})$.
More mathematically, it is good to develop the perspective, in probability, of not thinking in a totally "pointwise" fashion all the time. For one, lots of things are defined not pointwise but pointwise almost surely, so trying to stipulate "how many points land at a specific other point" is not likely to be well-defined. However, there is a special case where specifying a function "up to measure zero" actually gives you a unique pointwise function, namely if you also assume that the function has to be continuous. Likewise, there is a continuity assumption that allows you to think of the law of $\alpha$ as a pointwise function rather than a measure, namely the assumption that the law of $\alpha$ is absolutely continuous with respect to $\mu$, the underlying measure on $M$ - in this case you can express the law of $\alpha$ in terms of a distribution function, like so: $\mathbb{P}(A)= \int_A \rho(\theta) d\mu(\theta)$, for some measurable function $\rho$. However, it is emphatically not always the case that the law of $\alpha$ is absolutely continuous, even in physically realistic scenarios. Trying to think of the law of $\alpha$ off the bat in terms of a distribution function like $\rho$ is baking in an extraneous assumption to your problem, like assuming for no particular reason that the trajectory of a particle is differentiable.
Now, what the notion of "distribution" (in the sense of: Schwartz distribution, Dirac delta distribution) is doing is it relaxes our notion of function so that we can think of the law of $\alpha$ in terms of some such $\rho$, if we really insist. Even in this case, however, it is not always the case that a Schwartz distribution can be thought of as an "uncountable sum of deltas" as you seem to be trying to do. Therefore, it really is best practice to treat the law of $\alpha$ as the "primitive" object of study for your type of problem.
I hope that helps.
Addendum: now, you may ask, how do you actually compute the law of $\alpha$ in a reasonable way? The answer is that for some measurable functions, you can't, but the distribution function $\rho$ perspective also suffers from this problem. In reality, what you do is sample from $\alpha$ a (large) finite number of times, and hope that this procedure will allow you to approximate the law of $\alpha$ if you sample enough points (and you're not very unlucky). There are some general results in statistics concerning when/if this actually works; the keywords you should look for are that the sample gives you an empirical measure which converges in distribution to the law of $\alpha$.
Best Answer
Some hints:
(a) If $f\colon X\to\overline{\mathbb R}$ (with any $\sigma$-algebra $\mathcal B$) then for each $B\in\mathcal B$, $f^{-1}(B)$ is a subset of $X$.
(b) The sets which have measure $0$ are all the sets which doesn't contain $x$.
(c) Show that the only set which have counting measure $0$ is the emptyset.