Limit of a continuous function with a parameter Let $f(x,\alpha)$ be continuous function on $S=(0,1]\times[0,1]$. Suppose that for every segment $[\alpha,\alpha+\Delta\alpha]\in[0,1]$ there exists $x_0=x_0(\Delta \alpha)$ s.t. for $0<x<x_0$ on a vertical segment $I_{[\alpha,\alpha+\Delta\alpha]}(x)=\{(x,\beta),\alpha\le \beta\le \alpha+\Delta\alpha\}\ \ \ $ the estimate 
$$
\max_{\alpha\le \beta\le \alpha+\Delta\alpha}f(x,\beta)\ge g(x)\qquad\qquad{(1)}
$$
holds, where $\lim\limits_{x\to0+}g(x)=+\infty\,$.

Does it follow that $\lim\limits_{x\to0+}f(x,\alpha)=+\infty\ $ for all
  $\alpha$? Or may be a.e. on $[0,1]$?

It's straightforward to get from $(1)$ that for every $\alpha\in[0,1]$ there exists a sequence of points $(x_k,\alpha_k)\in S$ s.t. $(x_k,\alpha_k)\to(0,\alpha)$ and $\lim\limits_{k\to+\infty}f(x_k,\alpha_k)=+\infty$. But the sequence can approach $(0,\alpha)$  from the tangential direction and the desired result does not follow immediately.
 A: I shall construct an example with a dense set of exceptional $\alpha$'s.
Let $Q:=\ ]0,1]\times[0,1]$, and denote by $D$ the set of $\alpha\in[0,1]$ with a finite binary expansion. We begin by defining a "capacitor" consisting of two disjoint sets $A$ and $B$. Both  $A$ and $B$ are the union of disjoint horizontal  segments in $Q$. For each $$\alpha={2k-1\over2^n}\in D$$
the segment $\ ]0,2^{-n}]\times\{\alpha\}$ is put into $A$, and the segment $\ [2\cdot2^{-n},4\cdot2^{-n}]\times\{\alpha\}$ is put into $B$. In the following (incomplete!) figure $A$ is drawn in blue, $B$ in red.

Note that $d(p,A)+d(p,B)>0$ for all $p\in Q$. It follows that the "potential"
$$\phi(p):={d(p,A)\over d(p,A)+d(p,B)}\qquad\bigl(p=(x,\alpha)\in Q\bigr)$$
is well defined and continuous in $Q$; furthermore $\phi(p)=0$ when $p\in A$, and $\phi(p)=1$ when $p\in B$.
Consider now the function
$$f(x,\alpha):=\phi(x,\alpha)\cdot{1\over x}\qquad\bigl((x,\alpha)\in Q\bigr)\ .$$
This function is continuous in $Q$, and for all $\alpha\in D$ one has
$$\lim_{x\to 0+}f(x,\alpha)=0\ .$$
But it is easy to see that for $0<x<x_0(\Delta\alpha):=\Delta\alpha$ the following is guaranteed:
$$\max_{\alpha\leq\beta\leq\alpha+\Delta\alpha} f(x,\beta)={1\over x}\ .$$
A: Modifying the answer of @Christian Blatter, here is complete (negative) answer. Function $f(x)=g(x)\sin\frac{\alpha}x$ has the required property 
$$
\max_{\alpha\le \beta\le \alpha+\Delta\alpha}f(x,\beta)\ge g(x)
$$
for $x<\frac{\Delta\alpha}{2\pi}$, but $\liminf\limits_{x\to0}f(x,\alpha)=-\infty$ for all $\alpha\in(0,1]$. 
