Daniel Fischer gave a transparent explanation in terms of epigraphs $\{(x,y): y\ge f(x)\}$:
A function is convex if and only if its epigraph is convex, and the epigraph of a pointwise supremum is the intersection of the epigraphs. [Hence,] the pointwise supremum of convex functions is convex.
One can similarly argue from concavity, using the sets $\{(x,y): y\le f(x)\}$: this set is convex if and only if $f$ is concave. Taking infimum of functions results in taking the intersection of such sets.
In this case $I$ is an index set. For example, you could take $I = \{1,2,3\}$, $I = \mathbb{N}$, or even $I = \mathbb{R}$.
Your set of functions is then going to be $\{g_1, g_2, g_3\}$, $\{g_1,g_2, g_3, g_4, \dotsc\}$, or $\{g_i: i\in\mathbb{R}\}$ respectively (unfortunately there is no better way to write the last set since we can't count off real numbers).
From here, we define a new function $g$ as follows: fix an $x\in\mathbb{R}^n$. Compute all the $g_i$'s for this value of $x$ (for example, we may have $g_1(x) = 1$, $g_2(x) = 2$ and $g_3(x) = 3$). Then we take supremum of all these values, and let $g(x)$ denote that value (in this case, $3$). We then move on to another $x\in\mathbb{R}^n$, and in this manner we pick values of $g$ for all of $\mathbb{R}^n$. We need to take the supremum in our definition because the set of $g_i$'s evaluated at $x$ may be infinite. Like for the second $I$, I could have $\{g_i(x): i\in\mathbb{N}\} = \{0.9, 0.99, 0.999, 0.9999,\dotsc\}$. Here, we let $g(x) = 1$.
Edit: One more thing, since there is an important case I didn't address. I could also have my $g_i$'s infinitely growing, like $\{g_i(x): i\in\mathbb{N}\} = \{100, 200, 300, 400, \dotsc\}$. Based off the definition, we assign $g(x)$ the value $\infty$ if this happens.
Best Answer
Because the Lagrangian $L(x,\lambda,\mu)$ is affine in $\lambda$ and $\mu$, the Lagrange dual function $d(\lambda,\nu) = \inf_{x\in \mathcal{D}}L(x,\lambda,\nu)$ is always concave because it is the pointwise infimum of a set of affine functions, which is always concave. (You can also show that the supremum of a set of convex functions is convex.)