I'm looking for two (convergent) sequences of real-valued functions, $\{f_{k}\}$ and $\{g_{k}\}$, such that, for each $k$, $$\int_{-\infty}^{x}f_{k}(t)dt\leq \int_{-\infty}^{x}g_{k}(t)dt$$ at every $x$, and such that $$\int_{-\infty}^{x}f(t)dt> \int_{-\infty}^{x}g(t)dt$$ for some $x$. Moreover, I would like $f_{k}, g_{k}, f,$ and $g$ to be probability distribution functions. That is, they are monotone with an upper bound of $1$ and a lower bound of $0$.
These conditions may be too restrictive for such functions to exist, but I would appreciate the help. Thanks!
Edit: Moreover, I would like the distributions to be defined over a discrete space, if possible.
Using a dynkin-system argument together with the given condition $$\tag{1}\int_\infty^x f_k(t) dt \leq \int_{-\infty}^x g_k(t) dt$$ for every $x \in \mathbb{R}$, one can already conclude that this integral inequality holds not only for sets of type $(-\infty,x]$, but for all Borel-measurable sets. As a consequence one finds that $f_k(x) \le g_k(x)$ for almost every $x \in \mathbb{R}$.
If we have convergence in some of the usual used forms (i.e. $L^p$-convergence, in probability or (almost-everywhere) pointwise), then we get that a subsequence convergences almost-everywhere. Taking the limits and noting that the union of countable many nullsets is again a nullset, we conclude that $f(x) \le g(x)$ for almost every $x \in \mathbb{R}$. Integrating (that is possible, if - for example - $f$ and $g$ are probablity densites) gives $$\int_{-\infty}^x f(t) \ dt \le \int_{-\infty}^x g(t) \ dt$$ for every $x \in \mathbb{R}$. Thus, as you have already suggested, the conditions are too restrictive.
If (1) should only hold for one fixed $x \in \mathbb{R}$, then we first may note that the Fatou lemma implies that $$\int_{-\infty}^x f(t) dt \leq \liminf_{n \rightarrow \infty} \int_{-\infty}^x f_n(t) dt \le \liminf_{n \rightarrow \infty} \int_{-\infty}^x g_n(t) dt.$$ Thus, the requirements of the dominated convergence theorem have to be not fulfilled. For example take $f_n(x) = 1_{[0,1]}$ and $g_n = n 1_{(0,1/n]}$, then one has $$\int_{-\infty}^1 f_k(t) dt \le \int_{-\infty}^1 g_k(t) dt$$ and $f_n \rightarrow 1_{[0,1]}$ and $g_n \rightarrow 0$ almost everywhere. Thus $$0 \int_{-\infty}^1 g(t) dt < \int_{-\infty}^1 f(t) dt = 1.$$