Is the Set M Closed in the Space X?

[iris_m]

Homework Statement

Let $$X=(C([0,1]), || . ||_1 )$$, where $$||f||_1=\int_{0}^{1}|f(t)|dt$$.
Let $$M=\{f \in C([0,1]) : \int_{0}^{1}f(t)dt=2, f(1)=0\}$$.
Is M closed in X?

The Attempt at a Solution

I've tried the following:
Let $$f_n$$ be a sequence in M such that $$f_n \rightarrow f$$.
I'm checking whether $$f \in M$$, because that would prove that M is closed (if a set contains all the limits of its convergent sequences, it is closed).
There are obviously two conditions to check.
The first one:
$$\int_{0}^{1}f(t)dt=\int_{0}^{1}limf_n(t)dt=lim\int_{0}^{1}f_n(t)dt=lim 2=2$$.
Now I have to check the second one, that is, is $$f(1)=0$$, and I don't know how.

Any help is much appreciated.

 

[HallsofIvy]

If $f_n(1)= 0$ and $f_n\rightarrow f$, then... (What kind of convergence are you talking about? Pointwise? Uniform? In the norm?)

And is your justification for saying
$$\int_0^1 limf_n(t)dt= lim\int_0^1 f_n(t)dt$$?

 

[iris_m]

If $f_n(1)= 0$ and $f_n\rightarrow f$, then... (What kind of convergence are you talking about? Pointwise? Uniform? In the norm?)

In the norm.

And is your justification for saying
$$\int_0^1 limf_n(t)dt= lim\int_0^1 f_n(t)dt$$?

My justification would be that f_n are continuous functions, so integral and limit commute.