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How a thread can see stale reference of safely initialized object

I have been trying to figure out that how immutable objects which are safely published could be observed with stale reference.
public final class Helper {
private final int n;
public Helper(int n) {
this.n = n;
}
}
class Foo {
private Helper helper;
public Helper getHelper() {
return helper;
}
public void setHelper(int num) {
helper = new Helper(num);
}
}
So far I could understand that Helper is immutable and can be safely published. A reading thread either reads null or fully initialized Helper object as it won't be available until fully constructed. The solution is to put volatile in Foo class which I don't understand.

The fact that you are publishing a reference to an immutable object is irrelevant here.
If you are reading the value of a reference from multiple threads, you need to ensure that the write happens before a read if you care about all threads using the most up-to-date value.
Happens before is a precisely-defined term in the language spec, specifically the part about the Java Memory Model, which allows threads to make optimisations for example by not always updating things in main memory (which is slow), instead holding them in their local cache (which is much faster, but can lead to threads holding different values for the "same" variable). Happens-before is a relation that helps you to reason about how multiple threads interact when using these optimisations.
Unless you actually create a happens-before relationship, there is no guarantee that you will see the most recent value. In the code you have shown, there is no such relationship between writes and reads of helper, so your threads are not guaranteed to see "new" values of helper. They might, but they likely won't.
The easiest way to make sure that the write happens before the read would be to make the helper member variable final: the writes to values of final fields are guaranteed to happen before the end of the constructor, so all threads always see the correct value of the field (provided this wasn't leaked in the constructor).
Making it final isn't an option here, apparently, because you have a setter. So you have to employ some other mechanism.
Taking the code at face value, the simplest option would be to use a (final) AtomicInteger instead of the Helper class: writes to AtomicInteger are guaranteed to happen before subsequent reads. But I guess your actual helper class is probably more complicated.
So, you have to create that happens-before relationship yourself. Three mechanisms for this are:
Using AtomicReference<Helper>: this has similar semantics to AtomicInteger, but allows you to store a reference-typed value. (Thanks for pointing this out, #Thilo).
Making the field volatile: this guarantees visibility of the most recently-written value, because it causes writes to flush to main memory (as opposed to reading from a thread's cache), and reads to read from main memory. It effectively stops the JVM making this particular optimization.
Accessing the field in a synchronized block. The easiest thing to do would be to make the getter and setter methods synchronized. Significantly, you should not synchronize on helper, since this field is being changed.

Cite from Volatile vs Static in Java
This means that if two threads update a variable of the same Object concurrently, and the variable is not declared volatile, there could be a case in which one of the thread has in cache an old value.
Given your code, the following can happen:
Thread 1 calls getHelper() and gets null
Thread 2 calls getHelper() and gets null
Thread 1 calls setHelper(42)
Thread 2 calls setHelper(24)
And in this case your trouble starts regarding which Helper object will be used in which thread. The keyword volatile will at least solve the caching problem.

The variable helper is being read by multiple threads simultaneously. At the least, you have to make it volatile or the compiler will begin caching it in registers local to threads and any updates to the variable may not reflect in the main memory. Using volatile, when a thread starts reading a shared variable, it will clear its cache and fetch a fresh value from the global memory. When it finishes reading it, it will flush the contents of its cache into the main memory so that other threads may get the updated value.

Can a volatile variable that is never assigned null to ever contain null?

Can in the following conceptual Java example:
public class X implements Runnable {
public volatile Object x = new Object();
#Runnable
public void run() {
for (;;) {
Thread.sleep(1000);
x = new Object();
}
}
}
x ever be read as null from another thread?
Bonus: Do I need to declare it volatile (I do not really care about that value, it suffices that sometime in the future it will be the newly assigned value and never is null)

Technically, yes it can. That is the main reason for the original ConcurrentHashMap's readUnderLock. The javadoc even explains how:
Reads value field of an entry under lock. Called if value field ever appears to be null. This is possible only if a compiler happens to reorder a HashEntry initialization with its table assignment, which is legal under memory model but is not known to ever occur.
Since the HashEntry's value is volatile this type of reordering is legal on consturction.
Moral of the story is that all non-final initializations can race with object construction.
Edit:
#Nathan Hughes asked a valid question:
#John: in the OP's example wouldn't the construction have happened before the thread the runnable is passed into started? it would seem like that would impose a happens-before barrier subsequent to the field's initialization.
Doug Lea had a couple comments on this topic, the entire thread can be read here. He answered the comment:
But the issue is whether assignment of the new C instance to some other memory must occur after the volatile stores.
With the answer
Sorry for mis-remembering why I had treated this issue as basically settled:
Unless a JVM always pre-zeros memory (which usually not a good option), then
even if not explicitly initialized, volatile fields must be zeroed
in the constructor body, with a release fence before publication.
And so even though there are cases in which the JMM does not
strictly require mechanics preventing publication reordering
in constructors of classes with volatile fields, the only good
implementation choices for JVMs are either to use non-volatile writes
with a trailing release fence, or to perform each volatile write
with full fencing. Either way, there is no reordering with publication.
Unfortunately, programmers cannot rely on a spec to guarantee
it, at least until the JMM is revised.
And finished with:
Programmers do not expect that even though final fields are specifically
publication-safe, volatile fields are not always so.
For various implementation reasons, JVMs arrange that
volatile fields are publication safe anyway, at least in
cases we know about.
Actually updating the JMM/JLS to mandate this is not easy
(no small tweak that I know applies). But now is a good time
to be considering a full revision for JDK9.
In the mean time, it would make sense to further test
and validate JVMs as meeting this likely future spec.

This depends on how the X instance is published.
Suppose x is published unsafely, eg. through a non-volatile field
private X instance;
...
void someMethod() {
instance = new X();
}
Another thread accessing the instance field is allowed to see a reference value referring to an uninitialized X object (ie. where its constructor hasn't run yet). In such a case, its field x would have a value of null.
The above example translates to
temporaryReferenceOnStack = new memory for X // a reference to the instance
temporaryReferenceOnStack.<init> // call constructor
instance = temporaryReferenceOnStack;
But the language allows the following reordering
temporaryReferenceOnStack = new memory for X // a reference to the instance
instance = temporaryReferenceOnStack;
temporaryReferenceOnStack.<init> // call constructor
or directly
instance = new memory for X // a reference to the instance
instance.<init> // call constructor
In such a case, a thread is allowed to see the value of instance before the constructor is invoked to initialize the referenced object.
Now, how likely this is to happen in current JVMs? Eh, I couldn't come up with an MCVE.
Bonus: Do I need to declare it volatile (I do not really care about
that value, it suffices that sometime in the future it will be the
newly assigned value and never is null)
Publish the enclosing object safely. Or use a final AtomicReference field which you set.

No. The Java memory model guarantees that you will never seen x as null. x must always be the initial value it was assigned, or some subsequent value.
This actually works with any variable, not just volatile. What you are asking about is called "out of thin air values". C.f. Java Concurrency in Practice which talks about this concept in some length.
The other part of your question "Do I need to declare x as volatile:" given the context, yes, it should be either volatile or final. Either one provides safe publication for your object referenced by x. C.f. Safe Publication. Obviously, x can't be changed later if it's final.

Cost of using final fields

We know that making fields final is usually a good idea as we gain thread-safety and immutability which makes the code easier to reason about. I'm curious if there's an associated performance cost.
The Java Memory Model guarantees this final Field Semantics:
A thread that can only see a reference to an object after that object has been completely initialized is guaranteed to see the correctly initialized values for that object's final fields.
This means that for a class like this
class X {
X(int a) {
this.a = a;
}
final int a;
static X instance;
}
whenever Thread 1 creates an instance like this
X.instance = new X(43);
while (true) doSomethingEventuallyEvictingCache();
and Thread 2 sees it
while (X.instance == null) {
doSomethingEventuallyEvictingCache();
}
System.out.println(X.instance.a);
it must print 43. Without the final modifier, the JIT or the CPU could reorder the stores (first store X.instance and then set a=43) and Thread 2 could see the default-initialized value and print 0 instead.
When JIT sees final it obviously refrains from reordering. But it also has to force the CPU to obey the order. Is there an associated performance penalty?

Is there an associated performance penalty?
If you take a look at the source code of the JIT compiler, you will find the following comment regarding final member variables in the file src/share/vm/opto/parse1.cpp:
This method (which must be a constructor by the rules of Java) wrote a final. The effects of all initializations must be committed to memory before any code after the constructor publishes the reference to the newly constructor object. Rather than wait for the publication, we simply block the writes here. Rather than put a barrier on only those writes which are required to complete, we force all writes to complete.
The compiler emits additional instructions if there are final member variables. Most likely, these additional instructions cause a performance penalty. But it's unclear, if this impact is significant for any application.

How do final fields prevent other threads from seeing partially constructed objects?

I was looking into creating an immutable datatype that has final fields (including an array that is constructed and filled prior to being assigned to the final member field), and noticed that it seems that the JVM is specified to guarantee that any other thread that gets a reference to this object will see the initialized fields and array values (assuming no pointers to this are published within the constructor, see What is an "incompletely constructed object"? and How do JVM's implicit memory barriers behave when chaining constructors?).
I am curious how this is achieved without synchronizing every access to this object, or otherwise paying some significant performance penalty. According to my understanding, the JVM can achieve this by doing the following:
Issue a write-fence at the end of the constructor
Publish the reference to the new object only after the write-fence
Issue a read-fence any time you refer to a final field of an object
I can't think of a simpler or cheaper way of eliminating the risk of other threads seeing uninitialized final fields (or recursive references through final fields).
This seems like it could impose a severe performance penalty due to all of the read-fences in the other threads reading the object, but eliminating the read-fences introduces the possibility that the object reference is seen in another processor before it issues a read-fence or otherwise sees the updates to the memory locations corresponding to the newly initialized final fields.
Does anyone know how this works? And whether this introduces a significant performance penalty?

See the "Memory Barriers" section in this writeup.
A StoreStore barrier is required after final fields are set and before the object reference is assigned to another variable. This is the key piece of info you're asking about.
According to the "Reordering" section there, a store of a final field can not be reordered with respect to a store of a reference to the object containing the final field.
Additionally, it states that in v.afield = 1; x.finalField = v; ... ; sharedRef = x;, neither of the first two can be reordered with respect to the third; which ensures that stores to the fields of an object that is stored as a final field are themselves guaranteed to be visible to other threads before a reference to the object containing the final field is stored.
Together, this means that all stores to final fields must be visible to all threads before a reference to the object containing the field is stored.

Is this technically thread safe despite being mutable?

Yes, the private member variable bar should be final right? But actually, in this instance, it is an atomic operation to simply read the value of an int. So is this technically thread safe?
class Foo {
private int bar;
public Foo(int bar) {
this.bar = bar;
}
public int getBar() {
return bar;
}
}
// assume infinite number of threads repeatedly calling getBar on the same instance of Foo.
EDIT:
Assume that this is all of the code for the Foo class; any threads with a reference to a Foo instance will not be able to change the value of bar (without going to such lengths as using reflection etc.)

Final update: so my first conclusion happened to be right, just my reasoning was faulty :-( I re-edited my answer to make it somewhat coherent, not to hide the traces of my earlier blunder.
Conclusion
As #Wyzard pointed out, even though there is no way to change bar after construction, Foo is still not thread safe. The problem is not atomicity but visibility. If thread 1 is changing the value of bar in the constructor (from its default value of 0), there is no guarantee when other threads will get to see the new value (or whether they see it at all).
So foo looks like an immutable object. Quoting from Java Concurrency in Practice, section 3.4:
An object is immutable if:
Its state cannot be modified after construction;
All its fields are final; and
It is properly constructed (the this reference does not escape during construction).
Foo looks OK on 1) and 3), but not 2). And that is a crucial point, due to the reasoning above. Declaring a variable final is one way of ensuring its visibility between different threads. The other means are declaring bar volatile, or synchronizing its access method(s). But of course, in case of an immutable object, neither of these would make much sense.
Final Fields
So why do finalfields guarantee visibility? Answer from Java Concurrency in Practice, section 3.5.2:
Because immutable objects are so important, the JavaMemory Model offers a special guarantee of initialization safety for sharing immutable objects. As we've seen, that an object reference becomes visible to another thread does not necessarily mean that the state of that object is visible to the consuming thread. In order to guarantee a consistent view of the object's state, synchronization is needed.
Immutable objects, on the other hand, can be safely accessed even when synchronization is not used to publish the object reference. For this guarantee of initialization safety to hold, all of the requirements for immutability must be met: unmodi-fiable state, all fields are final, and proper construction. [...]
Immutable objects can be used safely by any thread without additional synchronization, even when synchronization is not used to publish them.
This guarantee extends to the values of all final fields of properly constructed objects; final fields can be safely accessed without additional synchronization. However, if final fields refer to mutable objects, synchronization is still required to access the state of the objects they refer to.
And what happens if the field is not final? Other threads may silently see a stale value of the field. There is no exception or any kind of warning - that is one reason why these kinds of bugs are so difficult to trace.