Vtyjkyuk

Which one of the following functions is left continuous and how to prove it?
$$
f(x;mathbfd,mathbfm) = x - sum_i=1^n(x-d_i)^+mathbf1_x<m_i,
$$
$$
g(x;mathbfd,mathbfm) = x - sum_i=1^n(x-d_i)^+mathbf1_xleq m_i,
$$
where, $mathbfb = (b_1,cdots,b_n) in mathbbR^n$, $mathbfm = (m_1,cdots,m_n) in mathbbR^n$ are given and $mathbf1_A = 1$ if $A$ holds, otherwise $0$.
The positive part function is $x^+= max(x,0)$.

Here, we only need to consider the left-continuity at $x = m_i, i = 1,cdots,n.$

By the definition of left continuity:
$$lim_xto x_0^- f(x) = f(x_0)$$

The function $f$ in your case has some value when it comes from the left, but suddenly jumps to another value as $x=m_i$.

Consider for example the biggest $m_i$, say $m_j$. The function $f$ jumps from some value (depending on $d_j$) to $m_j$, while $g$ does not.

So $g$ is left continuous, $f$ is not.

EDIT: even if the answer has been accepted, I'll give a hint for what the proof should look like.

For convenience, suppose the elements of $m$ are ordered like $m_1 < ... <m_n$. By linearity of the limit:
$$lim_xto m_1^- f(x) = m_1 - (m_1 - d_1)^+ \
f(m_1) = m_1$$
This fact alone proves that $f$ is not continuous from the left in general.
$$lim_xto m_1^- g(x) = m_1 - (m_1 - d_1)^+ \
g(m_1) = m_1- (m_1 - d_1)^+\
lim_xto m_2^- g(x) = m_2 - big( (m_2 - d_1)^+ + (m_2 - d_2)^+ big) \
g(m_2) = m_2 - big( (m_2 - d_1)^+ + (m_2 - d_2)^+ big)$$
And the same goes on for all other $m_i$. The key fact is that $1_A$ has a $leq$ rather than a $<$ in its $A$ proposition.